Calculus 2

Integration by parts

The product rule says how to differentiate f(x)g(x). Read backward, what does it say about integrating a product such as x\cos 5x, or a lone \ln x, that no substitution can reach?

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u-substitutionCalculus 1 Derivatives of logarithms and logarithmic differentiationCalculus 1

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Choosing an integration techniqueCalculus 2

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▴ 9 above Dividing before integrating Improper integrals with an infinite integrand

Integration by parts

Trigonometric integrals Integration by partial fractions ▾ Show all 16 (2 more)

Why this exists

Substitution reverses the chain rule, so it handles a composite function multiplied by the derivative of its inside. Many integrands do not have that shape. In \int x\cos 5x\,dx the factor x is not a constant times the derivative of 5x. In \int \ln x\,dx there is only one factor, and no basic rule has \ln x as a derivative.

The rule left to reverse is the product rule. Reversing it does not finish an integral in one stroke: it trades \int f g'\,dx for \int f' g\,dx, and the skill is to arrange the trade so the new integral is easier. That trade, made once, made repeatedly, or made twice in a loop that returns to the start, handles polynomials times exponentials or sines, logarithms, inverse trigonometric functions, and an exponential times a sine.

Definition

Day 1: the formula and the choice of parts

Let u = f(x) and v = g(x) have continuous derivatives on an interval, and write du = f'(x)\,dx and dv = g'(x)\,dx. Integration by parts is the identity

\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx,

written shortly as \int u\,dv = uv - \int v\,du. On [a, b] it reads \int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du.

To use it, the integrand is split into two parts: u, which will be differentiated, and dv, which will be integrated (the dx always goes with dv). The split is a choice, and a good one makes v findable and \int v\,du simpler than the integral we began with. So u is a factor that differentiation simplifies: a polynomial drops a degree, and \ln x, \arctan x and \arcsin x become the algebraic \displaystyle \frac{1}{x}, \displaystyle \frac{1}{1 + x^2} and \displaystyle \frac{1}{\sqrt{1 - x^2}} (Derivatives of logarithms and logarithmic differentiation). And dv is a factor whose antiderivative is known and no worse than itself: e^{kx}, \sin kx, \cos kx, a power of x, or plain dx. A logarithm or an inverse trigonometric function is nearly always u, since its antiderivative is the very thing being sought.

Day 2: repeating the step

Sometimes \int v\,du still needs parts. When u is a polynomial of degree n, each step lowers its degree, and the steps can be kept in a table: one column lists u and its derivatives down to 0, the other lists the function in dv and its successive antiderivatives. Each entry of the first column is multiplied by the entry one row down in the second, with signs +, -, +, -, \dots in turn. This bookkeeping is tabular integration. When neither factor dies out, as with e^{ax}\sin bx, two steps can bring back the original integral times a constant. The integral is then an unknown I in an equation, and we solve for it.

Intuition

Suppose u and v both increase as x runs from a to b, and plot the points (u, v). They trace a rising curve from (u_1, v_1) to (u_2, v_2), where u_1 = u(a), u_2 = u(b), and so on. The region between the curve and the u-axis has area \int v\,du; the region between the curve and the v-axis has area \int u\,dv. Together they fill the large rectangle, of area u_2v_2, with the small one, of area u_1v_1, cut away:

\int u\,dv + \int v\,du = u_2v_2 - u_1v_1 = \Big[uv\Big]_a^b.

Figure 1The two regions together fill the large rectangle with the small one removed, so \int u\,dv + \int v\,du = u_2v_2 - u_1v_1.

When one of the two areas is hard to find directly, the rectangles minus the other area give it. That is integration by parts.

Results

Integration by parts. If u and v have continuous derivatives on an interval, then on it

\int u\,dv = uv - \int v\,du,

and on [a, b],

\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du.

Tabular form. If p is a polynomial of degree n, G_1 is an antiderivative of g, and each G_{k+1} is an antiderivative of G_k, then

\begin{aligned} \int p\,g\,dx = {} & pG_1 - p'G_2 + p''G_3 - \dots \\ & + (-1)^n p^{(n)}G_{n+1} + C. \end{aligned}

Two antiderivatives found in Example 3: \int \ln x\,dx = x\ln x - x + C for x > 0, and \displaystyle \int \arctan x\,dx = x\arctan x - \frac{1}{2}\ln(1 + x^2) + C.

Why it is true

By the product rule, (uv)' = u'v + uv'. When u' and v' are continuous, so are both terms on the right; each has an antiderivative, and the integral of the sum is the sum of the integrals. Since uv is an antiderivative of the right side,

uv = \int u'v\,dx + \int uv'\,dx,

where each indefinite integral stands for a family of functions that differ by constants. Subtracting gives \int uv'\,dx = uv - \int u'v\,dx, which is \int u\,dv = uv - \int v\,du. For the definite form, the The fundamental theorem of calculus gives \int_a^b (u'v + uv')\,dx = \big[uv\big]_a^b; split the left side into two integrals and subtract.

Any antiderivative of dv serves as v. With v + K in its place, the right side becomes u(v + K) - \int (v + K)\,du, which is uv - \int v\,du + Ku - Ku plus a constant: the same family. So no constant is attached to v, and +\,C is written once, at the end.

For the tabular form, parts with u = p and v = G_1 gives \int pg\,dx = pG_1 - \int p'G_1\,dx. Parts again with u = p' and v = G_2 gives \int p'G_1\,dx = p'G_2 - \int p''G_2\,dx. Each step puts one more derivative on p, moves one place down the antiderivatives, and brings one more minus sign, so the signs alternate. After n + 1 steps the integral left over is \int p^{(n+1)}G_{n+1}\,dx = \int 0\,dx, a constant.

Worked examples

Example 1

Day 1: a polynomial times a cosine

Find \int x\cos 5x\,dx.

Solution

Take u = x, which differentiation turns into 1, and dv = \cos 5x\,dx, so du = dx and \displaystyle v = \frac{1}{5}\sin 5x:

\begin{aligned} \int x\cos 5x\,dx &= \frac{x}{5}\sin 5x - \frac{1}{5}\int \sin 5x\,dx \\ &= \frac{x}{5}\sin 5x + \frac{1}{25}\cos 5x + C. \end{aligned}

Check: the derivative is \displaystyle \frac{1}{5}\sin 5x + x\cos 5x - \frac{5}{25}\sin 5x = x\cos 5x.

The other split, u = \cos 5x and dv = x\,dx, is legal but goes the wrong way. Then \displaystyle v = \frac{x^2}{2}, du = -5\sin 5x\,dx, and the formula gives \displaystyle \frac{x^2}{2}\cos 5x + \frac{5}{2}\int x^2\sin 5x\,dx, with the power of x raised. The formula holds for every split; only some splits make progress.

Example 2

Day 1: a logarithm times a power

Find \int x^4\ln x\,dx, and then \int_1^e x^4\ln x\,dx.

Solution

Taking dv = \ln x\,dx would need an antiderivative of \ln x, which we do not yet have. So u = \ln x and dv = x^4\,dx, giving \displaystyle du = \frac{1}{x}\,dx and \displaystyle v = \frac{x^5}{5}. The logarithm drops out of the new integral:

\begin{aligned} \int x^4\ln x\,dx &= \frac{x^5}{5}\ln x - \int \frac{x^5}{5}\cdot\frac{1}{x}\,dx \\ &= \frac{x^5}{5}\ln x - \frac{x^5}{25} + C, \quad x > 0. \end{aligned}

For the definite integral, \ln e = 1 and \ln 1 = 0:

\begin{aligned} \int_1^e x^4\ln x\,dx &= \left(\frac{e^5}{5} - \frac{e^5}{25}\right) - \left(0 - \frac{1}{25}\right) \\ &= \frac{4e^5 + 1}{25} \approx 23.79. \end{aligned}

Example 3

Day 1: a single factor, ln x and arctan x

Find \int \ln x\,dx and \int \arctan x\,dx.

Solution

Neither looks like a product, but each is a product with 1. Take u = \ln x and dv = dx, so \displaystyle du = \frac{1}{x}\,dx and v = x. For x > 0,

\begin{aligned} \int \ln x\,dx &= x\ln x - \int x\cdot\frac{1}{x}\,dx \\ &= x\ln x - x + C. \end{aligned}

The same choice works for \arctan x: u = \arctan x, \displaystyle du = \frac{1}{1 + x^2}\,dx, v = x. The new integral yields to the substitution w = 1 + x^2, dw = 2x\,dx:

\begin{aligned} \int \arctan x\,dx &= x\arctan x - \int \frac{x}{1 + x^2}\,dx \\ &= x\arctan x - \frac{1}{2}\ln(1 + x^2) + C. \end{aligned}

No absolute value is needed, since 1 + x^2 > 0. Check: the derivative is \displaystyle \arctan x + \frac{x}{1 + x^2} - \frac{x}{1 + x^2} = \arctan x.

Example 4

Day 2: a cubic times an exponential, in a table

Find \int x^3 e^{2x}\,dx.

Solution

With u = x^3 each step of parts lowers the power by one, so the table is short. The middle column differentiates x^3 down to 0; the right column integrates e^{2x} again and again, each time multiplying by \displaystyle \frac{1}{2}.

sign	x^3 and its derivatives	e^{2x} and its antiderivatives
+	x^3	e^{2x}
-	3x^2	\displaystyle \frac{1}{2}e^{2x}
+	6x	\displaystyle \frac{1}{4}e^{2x}
-	6	\displaystyle \frac{1}{8}e^{2x}
	0	\displaystyle \frac{1}{16}e^{2x}

Each middle entry times the right entry one row down, with its sign:

\begin{aligned} \int x^3 e^{2x}\,dx &= \frac{x^3}{2}e^{2x} - \frac{3x^2}{4}e^{2x} \\ &\quad + \frac{6x}{8}e^{2x} - \frac{6}{16}e^{2x} + C \\ &= \frac{e^{2x}}{8}\left(4x^3 - 6x^2 + 6x - 3\right) + C. \end{aligned}

Each row is one integration by parts. The first gives \displaystyle \frac{x^3}{2}e^{2x} - \int \frac{3x^2}{2}e^{2x}\,dx, and the minus sign in front of the leftover integral is why the signs alternate. Check: the derivative is \displaystyle \frac{e^{2x}}{4}(4x^3 - 6x^2 + 6x - 3) \displaystyle + \frac{e^{2x}}{8}(12x^2 - 12x + 6), and everything but x^3e^{2x} cancels.

Example 5

Day 2: an integral that comes back

Find \int e^{2x}\sin 3x\,dx.

Solution

Call the integral I. Neither factor simplifies when differentiated, so a table would never end. Take u = \sin 3x and dv = e^{2x}\,dx, so du = 3\cos 3x\,dx and \displaystyle v = \frac{1}{2}e^{2x}:

I = \frac{1}{2}e^{2x}\sin 3x - \frac{3}{2}\int e^{2x}\cos 3x\,dx.

Apply parts to the new integral with the same kind of split, u = \cos 3x and dv = e^{2x}\,dx, so du = -3\sin 3x\,dx:

\int e^{2x}\cos 3x\,dx = \frac{1}{2}e^{2x}\cos 3x + \frac{3}{2}I.

Keeping the exponential as dv matters: taking dv = \cos 3x\,dx here would undo the first step and return I = I, true and empty. Substituting,

I = \frac{1}{2}e^{2x}\sin 3x - \frac{3}{4}e^{2x}\cos 3x - \frac{9}{4}I.

Adding \displaystyle \frac{9}{4}I to both sides gives \displaystyle \frac{13}{4}I = \frac{e^{2x}}{4}(2\sin 3x - 3\cos 3x), so

\int e^{2x}\sin 3x\,dx = \frac{e^{2x}}{13}\left(2\sin 3x - 3\cos 3x\right) + C.

The equation was solved for one antiderivative; every other differs from it by a constant, which is why +\,C appears only now. Check: the derivative is \displaystyle \frac{e^{2x}}{13} times 4\sin 3x - 6\cos 3x + 6\cos 3x + 9\sin 3x, which is e^{2x}\sin 3x.

Example 6

Day 2: total exposure to a drug

In Oysha’s pharmacokinetic study, the drug’s concentration in a patient’s blood t hours after a dose is modeled by C(t) = 6t\,e^{-t/2} milligrams per liter. The total exposure over the first 12 hours is the area \int_0^{12} 6t\,e^{-t/2}\,dt, in milligram-hours per liter. Find it and the average concentration over those hours. Dilfuza asks how much exposure the model would add if it ran forever.

Solution

Take u = 6t and dv = e^{-t/2}\,dt, so du = 6\,dt and v = -2e^{-t/2}. The definite form gives

\begin{aligned} \int_0^{12} 6t\,e^{-t/2}\,dt &= \Big[-12t\,e^{-t/2}\Big]_0^{12} \\ &\quad + 12\int_0^{12} e^{-t/2}\,dt \\ &= -144e^{-6} + 12\Big[-2e^{-t/2}\Big]_0^{12} \\ &= -144e^{-6} + 24\left(1 - e^{-6}\right) \\ &= 24 - 168e^{-6} \approx 23.58. \end{aligned}

Figure 2Oysha’s concentration model (Example 6). The shaded area is the total exposure over the first 12 hours; the peak is \displaystyle \frac{12}{e} \approx 4.41 milligrams per liter at t = 2.

The exposure is about 23.58 milligram-hours per liter, and the average concentration is \displaystyle \frac{1}{12}\left(24 - 168e^{-6}\right) \approx 1.97 milligrams per liter.

For Dilfuza’s question, the same work up to T hours, with antiderivative -(12t + 24)e^{-t/2}, gives \int_0^T C(t)\,dt = 24 - (12T + 24)e^{-T/2}. As T grows, (12T + 24)e^{-T/2} \to 0, since the exponential decay outruns the linear growth (Comparing growth rates at infinity). So the exposure approaches 24, and the first 12 hours carry \displaystyle \frac{23.58}{24}, about 98.3\% of it. Only about 0.42 more would ever be added.

Example 7

Day 2: a reduction formula for powers of ln x

Shahnoza needs \int_1^e (\ln x)^3\,dx. Find a formula linking J_n = \int_1^e (\ln x)^n\,dx to J_{n-1}, and use it.

Solution

As in Example 3, the logarithm is u and dv = dx: u = (\ln x)^n, \displaystyle du = n(\ln x)^{n-1}\cdot\frac{1}{x}\,dx, v = x. The x in v cancels the \displaystyle \frac{1}{x} in du. For n \ge 1,

\begin{aligned} J_n &= \Big[x(\ln x)^n\Big]_1^e - n\int_1^e (\ln x)^{n-1}\,dx \\ &= e - nJ_{n-1}, \end{aligned}

because the bracket is e \cdot 1^n - 1 \cdot 0^n = e. Start from J_0 = \int_1^e dx = e - 1:

\begin{aligned} J_1 &= e - (e - 1) = 1, \\ J_2 &= e - 2 \cdot 1 = e - 2, \\ J_3 &= e - 3(e - 2) = 6 - 2e. \end{aligned}

So \int_1^e (\ln x)^3\,dx = 6 - 2e \approx 0.563. This is plausible: on [1, e], 0 \le \ln x \le 1, so the integral lies between 0 and e - 1 \approx 1.72. A rule such as J_n = e - nJ_{n-1}, which lowers an exponent by one each time it is used, is a reduction formula: three integrations by parts packed into one line.

Boundaries

Continuous derivatives on the whole interval. The derivation needs u' and v' continuous where the integral is taken; the definite form needs it on all of [a, b]. An answer also carries the domain of its functions: \int \ln x\,dx = x\ln x - x + C holds only for x > 0, and an answer containing \arcsin x only for -1 < x < 1.
A trade, not a guarantee. The formula holds for every split but helps only when \int v\,du is simpler. Some integrals, such as \displaystyle \int \frac{e^x}{x}\,dx and \int e^{x^2}\,dx, have no antiderivative built from the usual functions (a fact stated here without proof), and no split finds one. With \displaystyle u = \frac{1}{x} and dv = e^x\,dx the first becomes \displaystyle \frac{e^x}{x} + \int \frac{e^x}{x^2}\,dx, which is worse.
When a table ends. A table ends only when the differentiated column reaches 0, so that column must hold a polynomial, and the other factor must be integrable again and again. The derivatives of \ln x, namely \displaystyle \frac{1}{x}, -\frac{1}{x^2}, \dots, never reach 0; for a logarithm a single step, which removes it, is the right move. For e^{ax}\sin bx neither column ends, and the loop of Example 5 replaces the table.
Solving for the integral. The loop works when I returns as I = A - kI with k \ne -1, giving \displaystyle I = \frac{A}{1 + k}. If it returns as I = A + I, the equation says only that A is a constant, and nothing is learned; this happens when the second step undoes the first.

Where this goes next

Integration by parts runs through the rest of this unit. The odd powers of secant left over in Trigonometric integrals need parts: with u = \sec x and dv = \sec^2 x\,dx, \int \sec^3 x\,dx returns in the manner of Example 5 and is solved for. Trigonometric substitution and partial fractions take integrands that parts does not simplify; Choosing an integration technique reads an integrand to decide which method fits. Dilfuza’s limit in Example 6 is an improper integral, and there a parts step like Example 6’s is followed by a limit as the upper end grows without bound. Later, integrating by parts again and again on the error of a Taylor polynomial gives a formula for that error.

Exercises

Day 1: one step of parts

\int x e^{3x}\,dx
\displaystyle \int x\sin\frac{x}{2}\,dx
\int (2x + 1)e^{-x}\,dx
\displaystyle \int \sqrt{x}\,\ln x\,dx
\int \ln(x^2 + 1)\,dx
\int \arcsin x\,dx
\displaystyle \int \frac{\ln x}{x^2}\,dx
\int x\arctan 2x\,dx

Day 1: definite integrals

\int_0^{\pi} x\sin x\,dx
\int_1^e x^2\ln x\,dx
\int_0^1 x\,2^x\,dx

Day 1: reasoning

For \int x e^{3x}\,dx, carry out one step with u = e^{3x} and dv = x\,dx. Explain why this split makes no progress.
Redo Example 1 with \displaystyle v = \frac{1}{5}\sin 5x + 7 in place of \displaystyle v = \frac{1}{5}\sin 5x. Show that the final answer is the same family.

Day 2: repeated parts, tables and loops

\int x^2\sin 2x\,dx
\int x^4 e^{-x}\,dx
\int (x^2 - 3x)\cos x\,dx
\int e^{-x}\cos 2x\,dx
\int \sin(\ln x)\,dx
\int x^3 e^{x^2}\,dx (split off x\,e^{x^2}\,dx as dv)

Day 2: the drug study

Shahnoza tests a second drug whose concentration is C(t) = 8t\,e^{-t} milligrams per liter. Find its total exposure over the first 6 hours.
Dilfuza needs the average concentration in Oysha’s model, C(t) = 6t\,e^{-t/2}, over the first 4 hours.
In Oysha’s model, find \int_0^{12} t\,C(t)\,dt and divide it by the exposure from Example 6. The ratio is an average of the times t, each weighted by the concentration at that time. What does the ratio approach if 12 is replaced by a time T that grows without bound?

Reasoning

Substitute x = e^t in Shahnoza’s integral \int_1^e (\ln x)^3\,dx to turn it into an integral of t^3e^t. Evaluate that integral with a table and compare with Example 7.
Shahnoza integrates \displaystyle \int \frac{1}{x}\,dx by parts with \displaystyle u = \frac{1}{x} and dv = dx, gets \displaystyle \int \frac{1}{x}\,dx = 1 + \int \frac{1}{x}\,dx, subtracts, and concludes 0 = 1. Find the flaw.
Explain why the table for \int x^3e^{2x}\,dx ends but a table with \ln x in the differentiated column never does. Then find \int x^5\ln x\,dx with a single step.

Answers

\displaystyle \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C. u = x, \displaystyle v = \frac{1}{3}e^{3x}, leaving \displaystyle \frac{1}{3}\int e^{3x}\,dx.
\displaystyle -2x\cos\frac{x}{2} + 4\sin\frac{x}{2} + C. u = x, \displaystyle v = -2\cos\frac{x}{2}, leaving \displaystyle +2\int \cos\frac{x}{2}\,dx.
-(2x + 3)e^{-x} + C. u = 2x + 1, v = -e^{-x} give -(2x + 1)e^{-x} + 2\int e^{-x}\,dx.
\displaystyle \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C. u = \ln x, \displaystyle v = \frac{2}{3}x^{3/2}; the new integrand is \displaystyle \frac{2}{3}x^{1/2}.
x\ln(x^2 + 1) - 2x + 2\arctan x + C. u = \ln(x^2 + 1), v = x leave \displaystyle \int \frac{2x^2}{x^2 + 1}\,dx, and dividing first gives \displaystyle 2 - \frac{2}{x^2 + 1}.
\displaystyle x\arcsin x + \sqrt{1 - x^2} + C for -1 < x < 1. u = \arcsin x, v = x leave \displaystyle \int \frac{x}{\sqrt{1 - x^2}}\,dx = -\sqrt{1 - x^2}, by the substitution w = 1 - x^2.
\displaystyle -\frac{\ln x}{x} - \frac{1}{x} + C. u = \ln x, \displaystyle v = -\frac{1}{x}, leaving \displaystyle +\int \frac{1}{x^2}\,dx.
\displaystyle \frac{4x^2 + 1}{8}\arctan 2x - \frac{x}{4} + C. u = \arctan 2x, \displaystyle du = \frac{2}{1 + 4x^2}\,dx, \displaystyle v = \frac{x^2}{2} leave \displaystyle \int \frac{x^2}{1 + 4x^2}\,dx, and \displaystyle \frac{x^2}{1 + 4x^2} = \frac{1}{4} - \frac{1}{4(1 + 4x^2)} makes it \displaystyle \frac{x}{4} - \frac{1}{8}\arctan 2x.
\pi. u = x, v = -\cos x give \big[-x\cos x\big]_0^{\pi} = \pi plus \int_0^{\pi}\cos x\,dx = 0.
\displaystyle \frac{2e^3 + 1}{9} \approx 4.57. u = \ln x, \displaystyle v = \frac{x^3}{3}: \displaystyle \left[\frac{x^3}{3}\ln x - \frac{x^3}{9}\right]_1^e.
\displaystyle \frac{2}{\ln 2} - \frac{1}{(\ln 2)^2} \approx 0.804. u = x, \displaystyle v = \frac{2^x}{\ln 2} give \displaystyle \frac{2}{\ln 2} - \frac{1}{\ln 2}\int_0^1 2^x\,dx, and \displaystyle \int_0^1 2^x\,dx = \frac{1}{\ln 2}.
The integral becomes \displaystyle \frac{x^2}{2}e^{3x} - \frac{3}{2}\int x^2 e^{3x}\,dx. The new integrand has x^2 where the old had x, so the split moves away from an answer; u = x instead lowers the power to x^0.
The formula gives \displaystyle \frac{x}{5}\sin 5x + 7x minus \displaystyle \int\left(\frac{1}{5}\sin 5x + 7\right)dx, that is, \displaystyle \frac{x}{5}\sin 5x + 7x \displaystyle + \frac{1}{25}\cos 5x - 7x + C, which is Example 1’s answer: the 7x in uv cancels the 7x from \int v\,du.
\displaystyle -\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + C. Table: x^2, 2x, 2, 0 against \sin 2x, \displaystyle -\frac{1}{2}\cos 2x, \displaystyle -\frac{1}{4}\sin 2x, \displaystyle \frac{1}{8}\cos 2x.
-e^{-x}\left(x^4 + 4x^3 + 12x^2 + 24x + 24\right) + C. Table: x^4, 4x^3, 12x^2, 24x, 24 against -e^{-x}, e^{-x}, -e^{-x}, \dots; each sign combines with the alternating antiderivative to leave -e^{-x} on every term.
(x^2 - 3x - 2)\sin x + (2x - 3)\cos x + C. Table: x^2 - 3x, 2x - 3, 2 against \sin x, -\cos x, -\sin x.
\displaystyle \frac{e^{-x}}{5}(2\sin 2x - \cos 2x) + C. Two steps with dv = e^{-x}\,dx give I = -e^{-x}\cos 2x + 2e^{-x}\sin 2x - 4I, so 5I = e^{-x}(2\sin 2x - \cos 2x).
\displaystyle \frac{x}{2}\big(\sin(\ln x) - \cos(\ln x)\big) + C for x > 0. u = \sin(\ln x), v = x give I = x\sin(\ln x) - \int \cos(\ln x)\,dx, and the same move gives \int \cos(\ln x)\,dx = x\cos(\ln x) + I; so 2I = x\sin(\ln x) - x\cos(\ln x).
\displaystyle \frac{1}{2}e^{x^2}(x^2 - 1) + C. u = x^2, dv = x\,e^{x^2}\,dx, so \displaystyle v = \frac{1}{2}e^{x^2} by substitution, and \displaystyle \int v\,du = \int x\,e^{x^2}\,dx = \frac{1}{2}e^{x^2}.
8 - 56e^{-6} \approx 7.86 milligram-hours per liter. u = 8t, v = -e^{-t} give the antiderivative -8(t + 1)e^{-t}, which is -56e^{-6} at 6 and -8 at 0.
6 - 18e^{-2} \approx 3.56 milligrams per liter. With the antiderivative -(12t + 24)e^{-t/2} from Example 6, \int_0^4 C(t)\,dt = 24 - 72e^{-2} \approx 14.26; divide by 4.
96 - 2400e^{-6} \approx 90.05, and the ratio is about 3.82 hours; it approaches 4 hours. Table: 6t^2, 12t, 12, 0 against e^{-t/2}, -2e^{-t/2}, 4e^{-t/2}, -8e^{-t/2} gives the antiderivative -(12t^2 + 48t + 96)e^{-t/2}. Up to T the integral is 96 - (12T^2 + 48T + 96)e^{-T/2}, which tends to 96, and the exposure tends to 24, so the ratio tends to \displaystyle \frac{96}{24} = 4.
6 - 2e, as in Example 7. With x = e^t: dx = e^t\,dt, \ln x = t, and x = 1, e become t = 0, 1, so the integral is \int_0^1 t^3e^t\,dt. The table gives e^t\left(t^3 - 3t^2 + 6t - 6\right), which is -2e at 1 and -6 at 0.
The step itself is right: \displaystyle du = -\frac{1}{x^2}\,dx and v = x give \displaystyle \int \frac{1}{x}\,dx = 1 + \int \frac{1}{x}\,dx. But each side is a family of antiderivatives, \ln\lvert x \rvert + C_1 and 1 + \ln\lvert x \rvert + C_2, and these are the same family (take C_1 = C_2 + 1). Subtracting treats \displaystyle \int \frac{1}{x}\,dx as one function, which it is not. With limits, \big[1\big]_a^b = 0 and no contradiction appears.
\displaystyle \frac{x^6}{6}\ln x - \frac{x^6}{36} + C. Each derivative of x^3 lowers the degree, so the fourth is 0 and the last leftover integral is \int 0\,dx. Derivatives of \ln x are \displaystyle \frac{1}{x}, -\frac{1}{x^2}, \frac{2}{x^3}, \dots, never 0. But one step with u = \ln x, \displaystyle v = \frac{x^6}{6} already leaves \displaystyle \int \frac{x^5}{6}\,dx, a power with no logarithm.