Choosing an integration technique
Every method so far was taught with its own integrals in front of it. Faced with an integral that comes with no label, how do you tell which method it needs, and what happens when none of them can finish it?
Where you are 16 in Integrals
When you reach for it
Each method lesson so far began by announcing its method: u-substitution, integration by parts, Trigonometric integrals, Trigonometric substitution, partial fractions, with division and completing the square as preparation. An integral met in the wild carries no such label. It may also yield to two methods at very different cost, need a rewrite before any method fits, or need one method to set up another.
This lesson is about the reading, not the methods: looking at the shape of an integrand, choosing a route, and comparing routes. Day 2 adds the hard fact behind all of it. Some integrands have no antiderivative that can be written as a formula, and no amount of cleverness will produce one. Then the question changes from “what is the formula” to “what do we do instead”.
The procedure
Day 1: reading the shape, and rewriting first
Rewrite before choosing. Many integrals become basic forms after algebra: expanding a product, splitting a fraction over its denominator, dividing a top-heavy rational function, canceling a common factor, or applying an identity such as \sin^2 x + \cos^2 x = 1 or 1 + \cos 2x = 2\cos^2 x. Each rewrite replaces the integrand by an equal function on the interval in question, so it changes nothing about the integral. It is always worth a look first, because when it works the integral is finished before a method has started.
Then read the shape. The table pairs what an integrand shows with the move that shape calls for. It is a summary of what the earlier lessons proved, not a ranking: several rows can fit one integral.
| The integrand shows | First move to try |
|---|---|
| a function and its derivative as a factor, up to a constant | u-substitution |
| a product of two unlike kinds (power and exponential, power and trig, a log or inverse trig with a power) | parts, differentiating the factor that simplifies |
| powers of \sin and \cos, or of \sec and \tan | the identities of trigonometric integrals |
| \displaystyle \sqrt{a^2 - x^2}, \displaystyle \sqrt{a^2 + x^2}, \displaystyle \sqrt{x^2 - a^2} with no partner factor | trigonometric substitution |
| a rational function | divide if top-heavy, factor the bottom, then partial fractions or completing the square |
| a root of a linear expression, \displaystyle e^{\sqrt{x}}, or \ln x inside something else | a substitution that removes it, then whatever the new integral needs |
The first row deserves first look, because it is the cheapest. A root such as \displaystyle \sqrt{9 - x^2} points to trigonometric substitution, but if an odd power of x sits beside it, the derivative of the inside is present and a plain substitution is shorter (Example 2). A rational function can always be done by partial fractions, but if its numerator is the derivative of its denominator, the answer is one logarithm (Example 1).
Why one route can be much longer. Two correct routes must give answers that differ by a constant on each interval, since two antiderivatives of one function on an interval differ by a constant. They need not cost the same. The shorter route is usually the one that uses more of the integrand’s structure: a substitution that consumes a whole factor, or an identity that collapses a fraction.
Substitutions that unlock another method. A substitution need not finish an integral; it may only change its kind. With \displaystyle t = \sqrt{x}, an integrand built from \displaystyle \sqrt{x} becomes a function of t, and dx = 2t\,dt adds a power of t. The new integral may be a product for parts (Example 4) or a rational function for division or partial fractions (Example 5). The substitution is legitimate because x = t^2 is one-to-one for t \ge 0, with a continuous derivative 2t that is nonzero for t > 0, so the inverse substitution theorem of Trigonometric substitution applies.
Check by differentiating. Whatever the route, differentiating the answer must return the integrand. This is the only check that does not depend on the route taken.
Day 2: chains, and integrands with no elementary antiderivative
Elementary functions. A function is elementary if it can be built from constants and x by finitely many additions, subtractions, multiplications, divisions, roots, and compositions with e^x, \ln x, the trig functions and the inverse trig functions. Every antiderivative in the method lessons was elementary.
Some antiderivatives are not. The fundamental theorem says every function f continuous on an interval has an antiderivative there, namely F(x) = \int_a^x f(t)\,dt (see The first fundamental theorem of calculus). What can fail is that this F is elementary. In papers of the 1830s Joseph Liouville showed that, for many integrands, it never is. One consequence, stated here without proof, covers integrands of the form f(x)\,e^{g(x)}.
Theorem (Liouville’s criterion, stated without proof). Let f and g be rational functions with g not constant. Then \int f(x)\,e^{g(x)}\,dx is elementary exactly when there is a rational function R with R' + g'R = f, and in that case the integral is R(x)\,e^{g(x)} + C.
For \displaystyle \frac{e^x}{x} (with \displaystyle f = \frac{1}{x}, g = x) the criterion asks for a rational R with \displaystyle R' + R = \frac{1}{x}. If R has a pole of order k \ge 1 at 0, then R' has one of order k + 1, which R cannot cancel, so the left side has a pole of order at least 2 while the right has order 1. If R has no pole at 0, the left side is finite there and the right side is not. No such R exists, so \displaystyle \int \frac{e^x}{x}\,dx is not elementary, as Integration by parts stated. Results of the same kind, also stated without proof, show that \displaystyle \frac{\sin x}{x}, \displaystyle \frac{\cos x}{x}, \displaystyle \frac{1}{\ln x} and \cos(x^2) have no elementary antiderivatives, alongside the e^{-x^2} and \sin(x^2) named in earlier lessons.
Small changes matter. Being elementary is fragile. \displaystyle \int x\cos(x^2)\,dx = \frac{1}{2}\sin(x^2) + C is a one-line substitution, while \int \cos(x^2)\,dx is not elementary. A sum of two non-elementary pieces can be elementary, because their non-elementary parts can cancel (Example 6). A substitution can also move a non-elementary integral to another one: x = e^u turns \displaystyle \int \frac{dx}{\ln x} into \displaystyle \int \frac{e^u}{u}\,du.
What one does instead. Three tools remain, and each answers a different question.
- A numeric value. For a definite integral, the trapezoidal rule and Simpson’s rule give the number to any accuracy wanted.
- The accumulation function. F(x) = \int_a^x f(t)\,dt is a function with known derivative F' = f, so where it rises and falls, and where it peaks, can be read from f without any formula (Example 7).
- A series. Replacing the integrand by its power series and integrating term by term gives the antiderivative as a series, the subject of Representing functions as power series and Using series to evaluate limits and integrals.
Worked examples
Day 1: two routes, one far shorter
Find \int \dfrac{3x^2 + 2}{x^3 + 2x - 3}\,dx.
SolutionThe integrand is rational, which suggests partial fractions. But the derivative of the denominator is 3x^2 + 2, exactly the numerator, so u = x^3 + 2x - 3, du = (3x^2 + 2)\,dx gives
Partial fractions also works. The denominator has the root x = 1, so it factors as (x - 1)(x^2 + x + 3), and x^2 + x + 3 has discriminant 1 - 12 < 0, so it is irreducible. Setting
and clearing denominators gives 3x^2 + 2 = A(x^2 + x + 3) + (Bx + C)(x - 1). At x = 1, 5 = 5A, so A = 1. The x^2 coefficients give 3 = A + B, so B = 2, and the constants give 2 = 3A - C, so C = 1. The second piece is \displaystyle \frac{2x + 1}{x^2 + x + 3}, whose numerator is the derivative of its denominator, and
By the product law of logarithms this is \ln\lvert x^3 + 2x - 3 \rvert + C, the same family, valid on x < 1 and on x > 1. The substitution took one line because it used the one fact that mattered: the numerator is the derivative of the denominator.
Day 1: a root that looks like trigonometric substitution
Find \displaystyle \int x^3\sqrt{9 - x^2}\,dx.
SolutionThe root \displaystyle \sqrt{9 - x^2} fits x = 3\sin\theta, which would give 243\int \sin^3\theta\cos^2\theta\,d\theta, then an odd-power rewrite, then a triangle to return to x. But the integrand also holds an odd power of x, and x\,dx is a constant times the derivative of 9 - x^2. Let u = 9 - x^2, so du = -2x\,dx and x^2 = 9 - u. Writing x^3\,dx = x^2 \cdot x\,dx,
on -3 \le x \le 3. Check: with u = 9 - x^2 the derivative is \displaystyle \frac{1}{2}u^{3/2}(-2x) - \frac{9}{2}u^{1/2}(-2x), which is xu^{1/2}(9 - u), that is \displaystyle x^3\sqrt{9 - x^2}. The extra x^2 was not a problem, because it could be written in terms of u.
Day 1: an identity before anything else
Find \int \dfrac{dx}{1 + \sin x}.
SolutionNo factor is the derivative of 1 + \sin x, and the integrand is not a product or a power. Multiplying top and bottom by 1 - \sin x (allowed wherever \sin x \ne 1) uses 1 - \sin^2 x = \cos^2 x:
Both terms are basic derivatives read backward, so
Check: the derivative of \tan x - \sec x is \sec^2 x - \sec x\tan x, which equals \displaystyle \frac{1 - \sin x}{\cos^2 x} = \frac{1}{1 + \sin x}. The form \displaystyle \frac{\sin x - 1}{\cos x} tends to 0 where \sin x = 1, so the antiderivative extends across those points; it holds on each interval between zeros of 1 + \sin x.
Day 1: a substitution that sets up parts
Find \displaystyle \int e^{\sqrt{x}}\,dx.
SolutionNo factor \displaystyle \frac{1}{\sqrt{x}} is present, so \displaystyle u = \sqrt{x} does not finish it in one step. Still, \displaystyle \sqrt{x} is the awkward part, so remove it. Let \displaystyle t = \sqrt{x} for x \ge 0, so x = t^2 and dx = 2t\,dt:
This is a power times an exponential, the shape for parts. With u = t and dv = e^t\,dt, \int te^t\,dt = te^t - e^t + C, so
Check: by the product and chain rules the derivative is
Day 2: substitution, then division
Find \displaystyle \int \dfrac{dx}{1 + \sqrt[3]{x}}.
SolutionThe cube root is the obstacle, so let \displaystyle t = \sqrt[3]{x}, so x = t^3 and dx = 3t^2\,dt. The map t \mapsto t^3 is one-to-one on all real numbers, and its derivative 3t^2 is zero only at t = 0. The integral becomes \displaystyle 3\int \frac{t^2}{1 + t}\,dt, a top-heavy rational function. Dividing, t^2 = (t - 1)(t + 1) + 1, so \displaystyle \frac{t^2}{1 + t} = t - 1 + \frac{1}{1 + t} and
on x < -1 and on x > -1. Check in t: \displaystyle \frac{d}{dt} of the bracket is \displaystyle 3t - 3 + \frac{3}{1 + t} = \frac{3t^2}{1 + t}, and dividing by \displaystyle \frac{dx}{dt} = 3t^2 gives \displaystyle \frac{1}{1 + t} wherever t \ne 0. At x = 0 itself, where \displaystyle \frac{dx}{dt} = 0, the answer is continuous and its derivative tends to 1 from both sides, so by the mean value theorem its derivative there is 1 as well. Three ideas were chained: a substitution to clear the root, division to make the fraction proper, and a basic logarithm.
Day 2: two non-elementary pieces with an elementary sum
Pemba is asked for \int \dfrac{x\cos x - \sin x}{x^2}\,dx on x > 0. Splitting gives \displaystyle \int \frac{\cos x}{x}\,dx - \int \frac{\sin x}{x^2}\,dx, and neither piece is elementary. Find the integral.
SolutionApply parts to the second piece with u = \sin x and dv = x^{-2}\,dx, so du = \cos x\,dx and \displaystyle v = -\frac{1}{x}:
This shows the second piece is not elementary (it differs from the non-elementary \displaystyle \int \frac{\cos x}{x}\,dx by an elementary function), and it also shows the two cancel in the difference:
The quotient rule gives the same thing directly: \displaystyle \frac{d}{dx}\frac{\sin x}{x} = \frac{x\cos x - \sin x}{x^2}. The whole integrand was a derivative, and splitting it hid that. The same holds on x < 0. Being elementary is a property of the integrand as a whole, not of its pieces.
Day 2: Dolkar’s sign
Dolkar cuts a wooden sign for the teahouse that Pemba and Lhamo run. Its bottom edge lies on the x-axis and its top edge follows \displaystyle y = \frac{\sin x}{x} for 0 \le x \le \pi, with y = 1 at x = 0, in feet (Figure 1). (a) Estimate its area with Simpson’s rule on four subintervals. (b) Let \displaystyle S(x) = \int_0^x \frac{\sin t}{t}\,dt. Explain why the area is the largest value S takes on [0, 3\pi].
(a) \displaystyle \frac{\sin x}{x} has no elementary antiderivative, so a formula cannot be evaluated at the ends; a number can still be computed. The value 1 at x = 0 is the limit \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 from The squeeze theorem and the limit of sine x over x, so the integrand is continuous on [0, \pi] and the integral is an ordinary one. With \displaystyle h = \frac{\pi}{4} the nodes give f(0) = 1, \displaystyle f\left(\frac{\pi}{4}\right) = \frac{2\sqrt{2}}{\pi} \approx 0.9003, \displaystyle f\left(\frac{\pi}{2}\right) = \frac{2}{\pi} \approx 0.6366, \displaystyle f\left(\frac{3\pi}{4}\right) = \frac{2\sqrt{2}}{3\pi} \approx 0.3001 and f(\pi) = 0. Simpson’s rule gives
The sign is about 1.85 square feet. A finer rule gives 1.8519, so four subintervals were already accurate to three decimal places.
(b) By the fundamental theorem, \displaystyle S'(x) = \frac{\sin x}{x}, which is positive on (0, \pi) and (2\pi, 3\pi) and negative on (\pi, 2\pi). So S rises to a local maximum at \pi, falls to a local minimum at 2\pi and rises again (Figure 2). The area is S(\pi). The rise from 2\pi to 3\pi adds \displaystyle \int_{2\pi}^{3\pi} \frac{\sin t}{t}\,dt, and the fall from \pi to 2\pi removes \displaystyle \left\lvert \int_{\pi}^{2\pi} \frac{\sin t}{t}\,dt \right\rvert; the arches of \lvert \sin t \rvert have equal area, but the second is divided by larger t, so the rise is smaller than the fall and S(3\pi) < S(\pi). With S(0) = 0, the largest value on [0, 3\pi] is S(\pi). All of this came from the derivative, with no formula for S.
Boundaries
- “Not elementary” is a statement about formulas, not about existence. Every function continuous on an interval has an antiderivative there, the accumulation function; it is simply not a finite combination of the listed functions.
- Liouville’s criterion, as stated, needs f and g rational with g not constant. It says nothing about \displaystyle \int \sqrt{\sin x}\,dx, for example, which needs other results.
- A non-elementary indefinite integral can still have an exact definite value. For instance \displaystyle \int_0^{\infty} \frac{\sin x}{x}\,dx = \frac{\pi}{2}, stated here without proof; it is an improper integral and the value comes from methods beyond this course.
- A substitution such as \displaystyle t = \sqrt{x} must be one-to-one with a continuous derivative on the interval used, nonzero except possibly at isolated points that are then checked directly (as x = 0 is in Example 5), and every answer holds only on an interval where the integrand is defined: \displaystyle \ln\lvert 1 + \sqrt[3]{x} \rvert in Example 5 is two antiderivatives, one on each side of x = -1.
- The error bound for Simpson’s rule needs a bounded fourth derivative on the interval. For \displaystyle \frac{\sin x}{x}, filled in with 1 at 0, that holds.
- Routes differ in length, never in result: two correct antiderivatives on one interval differ by a constant, and if they do not, one of them is wrong.
Where this goes next
The question “which method” returns in every later integral: arc length produces roots like \displaystyle \sqrt{1 + f'(x)^2}, most of which are not elementary; volumes and work produce products and rational functions; separable differential equations reduce to two integrals that must each be chosen and done. When no elementary answer exists, Representing functions as power series turns \displaystyle \frac{\sin x}{x} into \displaystyle 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots and integrates it term by term, and Using series to evaluate limits and integrals uses that series to compute definite values with a guaranteed error.
Exercises
Day 1: choose a route and finish
- Find \int x\sec^2(x^2)\,dx.
- Find \int x^2 e^{-2x}\,dx.
- Find \int \dfrac{(x + 1)^2}{x}\,dx.
- Find \int \dfrac{\cos^2 x}{1 - \sin x}\,dx.
- Find \displaystyle \int x\sqrt{x^2 - 16}\,dx.
- Find \displaystyle \int \dfrac{\sqrt{1 - x^2}}{x^2}\,dx.
- Find \int \dfrac{dx}{1 + \cos 2x}.
Day 1: substitute, then another method
- Find \displaystyle \int \cos\sqrt{x}\,dx.
- Find \int \dfrac{e^x}{e^{2x} - 3e^x + 2}\,dx.
- Find \int \dfrac{dx}{x\left[(\ln x)^2 - 4\right]}.
- Find \int x^5 e^{x^3}\,dx.
Day 1: reasoning
- Find \int \sin x\cos x\,dx three ways: with u = \sin x, with u = \cos x, and with \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x. Show that the three answers differ by constants.
- Without evaluating, name the first move for each and say what in the integrand calls for it: (a) \int \dfrac{x^2}{x^3 - 5}\,dx; (b) \int \dfrac{dx}{x^2 - 3x - 10}; (c) \int \dfrac{dx}{x^2 - 2x + 5}; (d) \displaystyle \int \sqrt{x^2 - 2x + 5}\,dx; (e) \int x^3\arctan x\,dx.
Day 2: chains
- Find \displaystyle \int \dfrac{dx}{1 + \sqrt{x}}.
- Find \displaystyle \int \dfrac{\sqrt{x}}{1 + x}\,dx.
- Find \displaystyle \int \arctan\sqrt{x}\,dx.
Day 2: elementary or not
- Find \int e^x\left(\dfrac{1}{x} - \dfrac{1}{x^2}\right)dx, and name the rational R that Liouville’s criterion asks for.
- Find \int \left(\dfrac{1}{\ln x} - \dfrac{1}{(\ln x)^2}\right)dx on x > 1, by differentiating \dfrac{x}{\ln x}.
- Show that x = e^u turns \int \dfrac{dx}{\ln x} into \int \dfrac{e^u}{u}\,du, and that u = e^x turns \int e^{e^x}\,dx into the same integral. Explain why neither \displaystyle \int \frac{dx}{\ln x} nor \int e^{e^x}\,dx is elementary.
Day 2: numbers and accumulation functions
- \cos(x^2) has no elementary antiderivative. Estimate \int_0^1 \cos(x^2)\,dx with Simpson’s rule on four subintervals.
- Using \displaystyle \cos u = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \cdots, integrate the first three terms of the series for \cos(x^2) from 0 to 1 and compare with Exercise 20.
- Lhamo wants a narrower sign with top edge \displaystyle y = \frac{\sin x}{x} on \displaystyle 0 \le x \le \frac{\pi}{2}, in feet. Estimate its area with Simpson’s rule on two subintervals.
- For \displaystyle S(x) = \int_0^x \frac{\sin t}{t}\,dt as in Example 7, find S'(2\pi) and say whether S(4\pi) is larger or smaller than S(3\pi), with a reason.
Day 2: reasoning
- Pemba says “\displaystyle \frac{\sin x}{x} has no antiderivative.” Correct the statement and say what is true.
- Explain why \displaystyle \int_0^{\pi} \frac{\sin x}{x}\,dx is an ordinary definite integral and not an improper one, although \displaystyle \frac{\sin x}{x} is undefined at 0.
Answers
- \displaystyle \frac{1}{2}\tan(x^2) + C. With u = x^2, \displaystyle x\,dx = \frac{1}{2}\,du, the integral is \displaystyle \frac{1}{2}\int \sec^2 u\,du. It holds on each interval where \cos(x^2) \ne 0.
- \displaystyle -\frac{1}{4}e^{-2x}\left(2x^2 + 2x + 1\right) + C. Parts twice, differentiating the power: \displaystyle -\frac{1}{2}x^2e^{-2x} + \int xe^{-2x}\,dx, and \displaystyle \int xe^{-2x}\,dx = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C.
- \displaystyle \frac{x^2}{2} + 2x + \ln\lvert x \rvert + C. Expanding, \displaystyle \frac{x^2 + 2x + 1}{x} = x + 2 + \frac{1}{x}, on x < 0 and on x > 0.
- x - \cos x + C. Since \cos^2 x = (1 - \sin x)(1 + \sin x), the integrand is 1 + \sin x wherever \sin x \ne 1.
- \displaystyle \frac{1}{3}\left(x^2 - 16\right)^{3/2} + C, for \lvert x \rvert \ge 4. The root invites x = 4\sec\theta, but the factor x is half the derivative of x^2 - 16, so u = x^2 - 16 gives \displaystyle \frac{1}{2}\int u^{1/2}\,du.
- \displaystyle -\frac{\sqrt{1 - x^2}}{x} - \arcsin x + C, on (-1, 0) and on (0, 1). No partner factor, so x = \sin\theta: the integrand becomes \displaystyle \frac{\cos^2\theta}{\sin^2\theta} = \csc^2\theta - 1, giving -\cot\theta - \theta, and \displaystyle \cot\theta = \frac{\sqrt{1 - x^2}}{x}.
- \displaystyle \frac{1}{2}\tan x + C, on each interval where \cos x \ne 0. The identity 1 + \cos 2x = 2\cos^2 x makes the integrand \displaystyle \frac{1}{2}\sec^2 x.
- \displaystyle 2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + C. With \displaystyle t = \sqrt{x}, dx = 2t\,dt, the integral is 2\int t\cos t\,dt, and parts gives t\sin t + \cos t.
- \displaystyle \ln\left\lvert \frac{e^x - 2}{e^x - 1} \right\rvert + C, for x \ne 0 and x \ne \ln 2. With u = e^x, du = e^x\,dx, the integral is \displaystyle \int \frac{du}{(u - 1)(u - 2)} and \displaystyle \frac{1}{(u - 1)(u - 2)} = \frac{1}{u - 2} - \frac{1}{u - 1}.
- \displaystyle \frac{1}{4}\ln\left\lvert \frac{\ln x - 2}{\ln x + 2} \right\rvert + C, for x > 0, x \ne e^{\pm 2}. With u = \ln x, \displaystyle du = \frac{dx}{x}, the integral is \displaystyle \int \frac{du}{u^2 - 4} and \displaystyle \frac{1}{u^2 - 4} = \frac{1}{4}\left(\frac{1}{u - 2} - \frac{1}{u + 2}\right).
- \displaystyle \frac{1}{3}\left(x^3 - 1\right)e^{x^3} + C. With u = x^3, \displaystyle x^5\,dx = x^3 \cdot x^2\,dx = \frac{1}{3}u\,du, and \int ue^u\,du = (u - 1)e^u + C by parts.
- \displaystyle \frac{1}{2}\sin^2 x + C, \displaystyle -\frac{1}{2}\cos^2 x + C and \displaystyle -\frac{1}{4}\cos 2x + C. The first minus the second is \displaystyle \frac{1}{2}(\sin^2 x + \cos^2 x) = \frac{1}{2}; the first minus the third is \displaystyle \frac{1}{2}\sin^2 x + \frac{1}{4}\left(1 - 2\sin^2 x\right), which is \displaystyle \frac{1}{4}. Both differences are constants, as two antiderivatives of one function must be.
- (a) u = x^3 - 5: the numerator is a third of the derivative of the denominator. (b) Partial fractions: x^2 - 3x - 10 = (x - 5)(x + 2) factors into distinct linear factors. (c) Complete the square, (x - 1)^2 + 4, for an arctangent: the quadratic has no real roots. (d) Complete the square, then x - 1 = 2\tan\theta: a root of a sum of squares with no partner factor. (e) Parts with u = \arctan x, dv = x^3\,dx: differentiating \arctan x leaves a rational function, \displaystyle \frac{x^4}{4\left(1 + x^2\right)}, which then needs division.
- \displaystyle 2\sqrt{x} - 2\ln\left(1 + \sqrt{x}\right) + C. With \displaystyle t = \sqrt{x}, the integral is \displaystyle 2\int \frac{t}{1 + t}\,dt = 2\int \left(1 - \frac{1}{1 + t}\right)dt.
- \displaystyle 2\sqrt{x} - 2\arctan\sqrt{x} + C. With \displaystyle t = \sqrt{x}, the integral is \displaystyle 2\int \frac{t^2}{1 + t^2}\,dt = 2\int \left(1 - \frac{1}{1 + t^2}\right)dt.
- \displaystyle (x + 1)\arctan\sqrt{x} - \sqrt{x} + C. With \displaystyle t = \sqrt{x}, the integral is \int 2t\arctan t\,dt. Parts with u = \arctan t and v = t^2 + 1 (any antiderivative of 2t may be used) gives (t^2 + 1)\arctan t - \int dt.
- \displaystyle \frac{e^x}{x} + C, on x < 0 and on x > 0. Here \displaystyle f = \frac{1}{x} - \frac{1}{x^2} and g = x, and \displaystyle R = \frac{1}{x} satisfies \displaystyle R' + R = -\frac{1}{x^2} + \frac{1}{x} = f. The product rule confirms \displaystyle \frac{d}{dx}\frac{e^x}{x} = \frac{e^x}{x} - \frac{e^x}{x^2}.
- \displaystyle \frac{x}{\ln x} + C. By the quotient rule, \displaystyle \frac{d}{dx}\frac{x}{\ln x} = \frac{\ln x - 1}{(\ln x)^2} = \frac{1}{\ln x} - \frac{1}{(\ln x)^2}.
- With x = e^u, dx = e^u\,du and \ln x = u, so \displaystyle \int \frac{dx}{\ln x} = \int \frac{e^u}{u}\,du. With u = e^x, du = e^x\,dx, so \displaystyle dx = \frac{du}{u} and \displaystyle \int e^{e^x}\,dx = \int \frac{e^u}{u}\,du. If \displaystyle \int \frac{dx}{\ln x} were an elementary H(x), then H(e^u) would be an elementary antiderivative of \displaystyle \frac{e^u}{u}; if \int e^{e^x}\,dx were an elementary G(x), then G(\ln u) would be one. Both are impossible, since \displaystyle \int \frac{e^u}{u}\,du is not elementary.
- About 0.9045. With \displaystyle h = \frac{1}{4}, the estimate is \displaystyle \frac{1}{12} times \displaystyle \cos 0 + 4\cos\frac{1}{16} + 2\cos\frac{1}{4} \displaystyle + 4\cos\frac{9}{16} + \cos 1 \approx 10.8540. A finer rule gives 0.90452.
- \displaystyle \frac{977}{1080} \approx 0.9046. The series is \displaystyle \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots, and \int_0^1 of the first three terms is \displaystyle 1 - \frac{1}{10} + \frac{1}{216}. It agrees with Exercise 20 to three decimal places; the next term, \displaystyle -\frac{1}{9360}, is about -0.0001.
- About 1.371 square feet. With \displaystyle h = \frac{\pi}{4}, the estimate is \displaystyle \frac{\pi}{12}\left[1 + 4 \cdot \frac{2\sqrt{2}}{\pi} + \frac{2}{\pi}\right], which is \displaystyle \frac{\pi}{12} + \frac{2\sqrt{2}}{3} + \frac{1}{6} \approx 1.3713. A finer rule gives 1.3708.
- \displaystyle S'(2\pi) = \frac{\sin 2\pi}{2\pi} = 0. S(4\pi) is smaller: on (3\pi, 4\pi), \sin t < 0, so \displaystyle S'(x) = \frac{\sin x}{x} < 0 and S decreases there. Numerically S(3\pi) \approx 1.6748 and S(4\pi) \approx 1.4922.
- It has an antiderivative on every interval: with the value 1 at 0 the function is continuous, so \displaystyle S(x) = \int_0^x \frac{\sin t}{t}\,dt is an antiderivative on all real numbers by the fundamental theorem. What is true is that no antiderivative of it is an elementary function, so it cannot be written as a finite formula in the usual functions.
- Since \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1, filling in the value 1 at 0 makes the integrand continuous, hence bounded, on [0, \pi]. Changing a function at one point does not change a definite integral, so the integral is the ordinary integral of a continuous function. An improper integral needs an infinite interval or an integrand that is unbounded near a point, and neither happens here.