Calculus 1

Completing the square inside an integral

The arctangent form fits one over one plus x squared exactly. What do you do with one over x squared plus six x plus ten, which matches nothing in the table?

You need

Antiderivatives of trig, exponential and log functionsCalculus 1 Completing the squareAlgebra 1

Unlocks

Trigonometric substitutionCalculus 2

Where you are 16 in Integrals

Calculus8

Integrals16

▴ 6 above Improper integrals u-substitution

Completing the square inside an integral

Dividing before integrating Improper integrals with an infinite integrand ▾ Show all 16 (5 more)

When you reach for it

The arctangent and arcsine forms from Antiderivatives of trig, exponential and log functions fit \dfrac{1}{1 + x^2} and \displaystyle \dfrac{1}{\sqrt{1 - x^2}} exactly, and nothing else. An integrand such as \dfrac{1}{x^2 + 6x + 10} matches no line of that table, and splitting the quotient does not help. But its denominator is a parabola that never reaches zero, and completing the square writes it as (x + 3)^2 + 1: the denominator 1 + x^2 with x replaced by x + 3. The graph is the arctangent integrand slid 3 units left, and its antiderivative is \arctan x slid the same way.

Reach for this move when a quadratic with no real roots sits in a denominator, either alone or under a linear numerator, and when a quadratic that opens downward sits under a square root in a denominator.

The procedure

Reshaping the quadratic

Completing the square rewrites any quadratic with leading coefficient 1:

x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right).

Write \displaystyle h = -\frac{b}{2}. The constant \displaystyle c - \frac{b^2}{4} equals \displaystyle -\frac{1}{4}\left(b^2 - 4c\right), so it is positive exactly when the discriminant b^2 - 4c is negative, that is, when the quadratic has no real roots. Then it is a square k^2 with k > 0, and

x^2 + bx + c = (x - h)^2 + k^2.

A leading coefficient other than 1 is factored out of the denominator first and comes outside the integral as a constant. Under a square root the quadratic usually opens downward: factor out -1, complete the square inside, and distribute the -1 back, which gives k^2 - (x - h)^2.

The two shifted forms

For k > 0,

\int \frac{dx}{(x - h)^2 + k^2} = \frac{1}{k}\arctan\frac{x - h}{k} + C,

\int \frac{dx}{\sqrt{k^2 - (x - h)^2}} = \arcsin\frac{x - h}{k} + C,

the second on the interval h - k < x < h + k.

Both are checked by differentiating with the chain rule and the derivatives from Derivatives of inverse trigonometric functions. The inside function \displaystyle \frac{x - h}{k} has derivative \displaystyle \frac{1}{k}, so

\begin{aligned} &\frac{d}{dx}\left[\frac{1}{k}\arctan\frac{x - h}{k}\right] \\ &\quad = \frac{1}{k}\cdot\frac{1}{1 + \frac{(x - h)^2}{k^2}}\cdot\frac{1}{k} = \frac{1}{k^2 + (x - h)^2}, \end{aligned}

after multiplying the k^2 into the denominator. For the arcsine,

\begin{aligned} &\frac{d}{dx}\left[\arcsin\frac{x - h}{k}\right] \\ &\quad = \frac{1}{\sqrt{1 - \frac{(x - h)^2}{k^2}}}\cdot\frac{1}{k} = \frac{1}{\sqrt{k^2 - (x - h)^2}}. \end{aligned}

The last step moves k inside the root as \displaystyle \sqrt{k^2}, which equals k only because k > 0. The arcsine derivative exists only when \displaystyle -1 < \frac{x - h}{k} < 1, which is h - k < x < h + k, and that is also exactly where k^2 - (x - h)^2 is positive.

Figure 1 shows why the shift costs nothing. Replacing x by x - h slides a graph h units right, and every area under it slides along unchanged. Renaming x - h as a single letter u is the simplest case of u-substitution; here each answer is confirmed by differentiating instead.

Figure 1x^2 - 4x + 5 = (x - 2)^2 + 1, so y = \dfrac{1}{x^2 - 4x + 5} is y = \dfrac{1}{1 + x^2} moved 2 units right. Areas move with the graph: both shaded regions have area \displaystyle \frac{\pi}{4}.

A linear numerator

For \dfrac{px + q}{x^2 + bx + c}, compare the numerator with the derivative of the denominator, 2x + b. Any linear numerator splits as a multiple of 2x + b plus a constant:

px + q = \frac{p}{2}(2x + b) + \left(q - \frac{pb}{2}\right).

The first piece is \displaystyle \frac{p}{2} times \dfrac{2x + b}{x^2 + bx + c}, the derivative of the denominator over the denominator. By the chain rule (Derivatives of logarithms and logarithmic differentiation), \dfrac{d}{dx}\big[\ln Q(x)\big] = \dfrac{Q'(x)}{Q(x)} wherever Q(x) > 0, so that piece integrates to \displaystyle \frac{p}{2}\ln\left(x^2 + bx + c\right). No absolute value is needed, since (x - h)^2 + k^2 \ge k^2 > 0 for every x. The second piece is a constant times the arctangent form.

Worked examples

Example 1

The plain case

Find \int \dfrac{dx}{x^2 + 6x + 10}.

Solution

Half of 6 is 3 and 3^2 = 9, so x^2 + 6x + 10 = (x + 3)^2 + 1. Here h = -3 and k = 1, and the arctangent form gives

\int \frac{dx}{(x + 3)^2 + 1} = \arctan(x + 3) + C.

Check: \dfrac{d}{dx}\big[\arctan(x + 3)\big] = \dfrac{1}{1 + (x + 3)^2} \cdot 1, and 1 + (x + 3)^2 = x^2 + 6x + 10. The answer holds for every real x, because the denominator is never zero.

Example 2

A constant on top and a square that is not 1

Find \int \dfrac{5}{x^2 - 8x + 25}\,dx.

Solution

Half of -8 is -4, and (-4)^2 = 16, so x^2 - 8x + 25 = (x - 4)^2 + 9. The constant 5 comes out, and h = 4, k = 3:

\int \frac{5}{(x - 4)^2 + 9}\,dx = \frac{5}{3}\arctan\frac{x - 4}{3} + C.

Check: the derivative is \displaystyle \dfrac{5}{3} \cdot \dfrac{1}{1 + \frac{(x - 4)^2}{9}} \cdot \dfrac{1}{3}, which is \dfrac{5}{9 + (x - 4)^2}. Without the factor \displaystyle \frac{1}{k} in front, the derivative would come out 3 times too large.

Example 3

Hiwa and Sirwan, a fraction and a root

Find \int \dfrac{dx}{x^2 + 3x + 3}. Hiwa gets \displaystyle \dfrac{2}{\sqrt{3}}\arctan\dfrac{2x + 3}{\sqrt{3}} + C; Sirwan gets \displaystyle \dfrac{2\sqrt{3}}{3}\arctan\dfrac{\sqrt{3}(2x + 3)}{3} + C. Who is right?

Solution

Half of 3 is \displaystyle \frac{3}{2} and \displaystyle \left(\frac{3}{2}\right)^2 = \frac{9}{4}, so

x^2 + 3x + 3 = \left(x + \frac{3}{2}\right)^2 + \frac{3}{4}.

So \displaystyle h = -\frac{3}{2} and \displaystyle k = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}. Then \displaystyle \frac{1}{k} = \frac{2}{\sqrt{3}}, and \displaystyle \dfrac{x - h}{k} = \dfrac{x + \frac{3}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{2x + 3}{\sqrt{3}}, which is Hiwa’s answer. Check:

\begin{aligned} &\frac{d}{dx}\left[\frac{2}{\sqrt{3}}\arctan\frac{2x + 3}{\sqrt{3}}\right] \\ &\quad = \frac{2}{\sqrt{3}}\cdot\frac{2}{\sqrt{3}}\cdot\frac{1}{1 + \frac{(2x + 3)^2}{3}} \\ &\quad = \frac{4}{3 + (2x + 3)^2} = \frac{4}{4x^2 + 12x + 12}, \end{aligned}

which is \dfrac{1}{x^2 + 3x + 3}. Sirwan is right too: he rationalized, and \displaystyle \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}, so his function is Hiwa’s written another way, with the same constant C.

Example 4

A leading coefficient that is not 1

Find \int \dfrac{dx}{9x^2 - 6x + 5}.

Solution

Factor 9 out of the denominator, then complete the square inside. Half of \displaystyle -\frac{2}{3} is \displaystyle -\frac{1}{3}, whose square is \displaystyle \frac{1}{9}:

\begin{aligned} 9x^2 - 6x + 5 &= 9\left(x^2 - \frac{2}{3}x + \frac{5}{9}\right) \\ &= 9\left[\left(x - \frac{1}{3}\right)^2 + \frac{4}{9}\right]. \end{aligned}

So \displaystyle h = \frac{1}{3} and \displaystyle k = \frac{2}{3}, and the \displaystyle \frac{1}{9} comes outside:

\begin{aligned} \int \frac{dx}{9x^2 - 6x + 5} &= \frac{1}{9}\int \frac{dx}{\left(x - \frac{1}{3}\right)^2 + \frac{4}{9}} \\ &= \frac{1}{9}\cdot\frac{3}{2}\arctan\frac{x - \frac{1}{3}}{\frac{2}{3}} + C \\ &= \frac{1}{6}\arctan\frac{3x - 1}{2} + C. \end{aligned}

Since 9x^2 is the square of 3x, the same square appears faster as 9x^2 - 6x + 5 = (3x - 1)^2 + 4. Check either way: the derivative of \displaystyle \frac{1}{6}\arctan\frac{3x - 1}{2} is \displaystyle \dfrac{1}{6}\cdot\dfrac{3}{2}\cdot\dfrac{1}{1 + \frac{(3x - 1)^2}{4}}, which is \dfrac{1}{(3x - 1)^2 + 4}.

Example 5

A downward quadratic under a root

Find \displaystyle \int \dfrac{dx}{\sqrt{8 + 2x - x^2}} and say where the answer holds.

Solution

Factor out -1 and complete the square inside:

\begin{aligned} 8 + 2x - x^2 &= -\left(x^2 - 2x - 8\right) \\ &= -\left[(x - 1)^2 - 9\right] = 9 - (x - 1)^2. \end{aligned}

So k = 3 and h = 1, and the arcsine form gives

\int \frac{dx}{\sqrt{9 - (x - 1)^2}} = \arcsin\frac{x - 1}{3} + C,

on the interval -2 < x < 4. That interval is h - k < x < h + k. It is also where the integrand is defined: 8 + 2x - x^2 = (4 - x)(2 + x) is positive exactly between its roots -2 and 4. Check: the derivative of \displaystyle \arcsin\frac{x - 1}{3} is \displaystyle \dfrac{1}{3}\cdot\dfrac{1}{\sqrt{1 - \frac{(x - 1)^2}{9}}} = \dfrac{1}{\sqrt{9 - (x - 1)^2}}.

Example 6

Arez splits a linear numerator

Arez needs \int \dfrac{2x + 7}{x^2 + 2x + 5}\,dx.

Solution

The derivative of the denominator is 2x + 2, and 2x + 7 = (2x + 2) + 5. So the integral splits:

\begin{aligned} \int \frac{2x + 7}{x^2 + 2x + 5}\,dx &= \int \frac{2x + 2}{x^2 + 2x + 5}\,dx \\ &\quad + \int \frac{5}{(x + 1)^2 + 4}\,dx. \end{aligned}

The first is the derivative of the denominator over the denominator, so it gives \ln\left(x^2 + 2x + 5\right), with no absolute value because the denominator is always positive. The second has h = -1 and k = 2. Together,

\begin{aligned} \int \frac{2x + 7}{x^2 + 2x + 5}\,dx &= \ln\left(x^2 + 2x + 5\right) \\ &\quad + \frac{5}{2}\arctan\frac{x + 1}{2} + C. \end{aligned}

Check: the derivative is \displaystyle \dfrac{2x + 2}{x^2 + 2x + 5} + \dfrac{5}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{1 + \frac{(x + 1)^2}{4}}, and the second term is \dfrac{5}{4 + (x + 1)^2} = \dfrac{5}{x^2 + 2x + 5}. The two terms add to the integrand.

Example 7

A definite integral, and a quadratic that factors

(a) Evaluate \int_1^4 \dfrac{dx}{x^2 - 2x + 10}. (b) Sirwan tries the same move on \dfrac{1}{x^2 - 2x - 3}. What goes wrong?

Solution

(a) x^2 - 2x + 10 = (x - 1)^2 + 9, so \displaystyle \frac{1}{3}\arctan\frac{x - 1}{3} is an antiderivative on all of [1, 4], where the integrand is continuous. By the fundamental theorem,

\begin{aligned} \frac{1}{3}\arctan\frac{x - 1}{3}\Big|_1^4 &= \frac{1}{3}\arctan 1 - \frac{1}{3}\arctan 0 \\ &= \frac{\pi}{12}. \end{aligned}

(b) Completing the square gives x^2 - 2x - 3 = (x - 1)^2 - 4, a square minus 4 rather than plus. The arctangent form needs a plus: the derivative of \displaystyle \frac{1}{2}\arctan\frac{x - 1}{2} is \dfrac{1}{(x - 1)^2 + 4}, a different function. The discriminant 4 + 12 = 16 is positive, and indeed x^2 - 2x - 3 = (x - 3)(x + 1). Such an integrand has vertical asymptotes at x = 3 and x = -1, and its antiderivative is a combination of logarithms found by partial fractions, a later topic.

Boundaries

The arctangent form needs no real roots. Only when b^2 - 4c < 0 does completing the square give a square plus a positive k^2. When b^2 - 4c > 0 the quadratic factors and the integrand is split by partial fractions. When b^2 - 4c = 0 the quadratic is a perfect square (x - h)^2, and the power rule gives \displaystyle \int \frac{dx}{(x - h)^2} = -\frac{1}{x - h} + C on each side of h.
The arcsine form needs a positive constant minus a square, so a downward quadratic with two real roots, and it holds only strictly between those roots. A square root of (x - h)^2 + k^2 or of (x - h)^2 - k^2 is a different shape, handled by Trigonometric substitution.
The formulas use k > 0. A negative k flips the sign of the arcsine formula (Exercise 23).
A definite integral needs continuity on [a, b]. A denominator (x - h)^2 + k^2 is never zero, so the arctangent form works on any interval. The arcsine form works on [a, b] only if it sits strictly inside (h - k, h + k); at h \pm k the integrand is unbounded and the integral is improper.
The numerator must be of lower degree. A numerator of degree 2 or more is divided first, by Dividing before integrating.

Where this goes next

u-substitution names the shift u = x - h as one case of a general method. Partial fractions breaks a rational function into pieces, and every piece over an irreducible quadratic is integrated exactly as in Example 6, as a log part plus an arctangent part. Dividing before integrating often leaves such a piece as a remainder. Trigonometric substitution completes the square in the same way and then handles square roots of (x - h)^2 + k^2 and (x - h)^2 - k^2. With improper integrals, the arctangent form shows that the area under \dfrac{1}{(x - h)^2 + k^2} over the whole line is \dfrac{\pi}{k}.

Exercises

The arctangent form

Find \int \dfrac{dx}{x^2 + 4x + 5}.
Find \int \dfrac{dx}{x^2 - 6x + 13}.
Find \int \dfrac{3}{x^2 + 8x + 41}\,dx.
Find \int \dfrac{dx}{x^2 + 2x + 3}.
Find \int \dfrac{dx}{x^2 - 3x + 4}.
Find \int \dfrac{dt}{t^2 + 2t + 17}.

A leading coefficient first

Find \int \dfrac{dx}{4x^2 + 4x + 5}.
Find \int \dfrac{dx}{2x^2 - 4x + 4}.
Find \int \dfrac{dx}{3x^2 + 6x + 15}.

The arcsine form

Find \displaystyle \int \dfrac{dx}{\sqrt{4x - x^2}} and say where it holds.
Find \displaystyle \int \dfrac{dx}{\sqrt{-x^2 - 6x - 5}} and say where it holds.
Find \displaystyle \int \dfrac{dx}{\sqrt{5 + 4x - x^2}} and say where it holds.

A linear numerator

Find \int \dfrac{2x + 4}{x^2 + 4x + 13}\,dx.
Find \int \dfrac{x + 3}{x^2 + 2x + 10}\,dx.
Find \int \dfrac{3x - 1}{x^2 - 4x + 20}\,dx.
Find \int \dfrac{x}{x^2 + 6x + 18}\,dx.

Definite integrals

Evaluate \int_{-1}^{1} \dfrac{dx}{x^2 + 2x + 5}.
Evaluate \int_2^5 \dfrac{dx}{x^2 - 4x + 7}.
Evaluate \displaystyle \int_1^2 \dfrac{dx}{\sqrt{3 + 2x - x^2}}.
Evaluate \int_1^5 \dfrac{x + 1}{x^2 - 2x + 17}\,dx.

Reasoning

For each denominator, say whether \displaystyle \int \frac{dx}{Q(x)} is an arctangent form, and find the integral where the methods so far apply: (a) Q(x) = x^2 + 6x + 8, (b) Q(x) = x^2 + 6x + 9, (c) Q(x) = x^2 + 6x + 11.
Arez writes \int \dfrac{dx}{x^2 + 4x + 20} = \arctan\dfrac{x + 2}{4} + C. Find the error by differentiating, and correct it.
Hiwa says Example 5 could use k = -3 just as well, giving \arcsin\dfrac{x - 1}{-3} + C. Differentiate her function and explain what the condition k > 0 is for.
Explain why the log part of a linear-numerator integral needs no absolute value, and why the arctangent form holds for every real x.
Show that \int_h^{h + k} \dfrac{dx}{(x - h)^2 + k^2} = \dfrac{\pi}{4k} for every h and every k > 0, and use Figure 1 to explain why the answer does not depend on h.

Answers

\arctan(x + 2) + C. The denominator is (x + 2)^2 + 1.
\dfrac{1}{2}\arctan\dfrac{x - 3}{2} + C. The denominator is (x - 3)^2 + 4, so k = 2.
\dfrac{3}{5}\arctan\dfrac{x + 4}{5} + C. The denominator is (x + 4)^2 + 25.
\displaystyle \dfrac{1}{\sqrt{2}}\arctan\dfrac{x + 1}{\sqrt{2}} + C. The denominator is (x + 1)^2 + 2, so \displaystyle k = \sqrt{2}.
\displaystyle \dfrac{2}{\sqrt{7}}\arctan\dfrac{2x - 3}{\sqrt{7}} + C. The denominator is \displaystyle \left(x - \frac{3}{2}\right)^2 + \frac{7}{4}, so \displaystyle k = \frac{\sqrt{7}}{2} and \displaystyle \frac{x - 3/2}{k} = \frac{2x - 3}{\sqrt{7}}.
\dfrac{1}{4}\arctan\dfrac{t + 1}{4} + C. The denominator is (t + 1)^2 + 16.
\dfrac{1}{4}\arctan\dfrac{2x + 1}{2} + C. The denominator is (2x + 1)^2 + 4, and the derivative of the answer is \displaystyle \frac{1}{4}\cdot\frac{2}{2}\cdot\frac{4}{4 + (2x + 1)^2}, the integrand.
\dfrac{1}{2}\arctan(x - 1) + C. The denominator is 2\left[(x - 1)^2 + 1\right].
\dfrac{1}{6}\arctan\dfrac{x + 1}{2} + C. The denominator is 3\left[(x + 1)^2 + 4\right], so the answer is \displaystyle \frac{1}{3}\cdot\frac{1}{2}\arctan\frac{x + 1}{2}.
\arcsin\dfrac{x - 2}{2} + C on 0 < x < 4. Here 4x - x^2 = 4 - (x - 2)^2.
\arcsin\dfrac{x + 3}{2} + C on -5 < x < -1. Here -x^2 - 6x - 5 = -\left[(x + 3)^2 - 4\right], which is 4 - (x + 3)^2.
\arcsin\dfrac{x - 2}{3} + C on -1 < x < 5. Here 5 + 4x - x^2 = 9 - (x - 2)^2.
\ln\left(x^2 + 4x + 13\right) + C. The numerator is exactly the derivative of the denominator, so there is no arctangent part.
\dfrac{1}{2}\ln\left(x^2 + 2x + 10\right) + \dfrac{2}{3}\arctan\dfrac{x + 1}{3} + C. Split \displaystyle x + 3 = \frac{1}{2}(2x + 2) + 2; the denominator is (x + 1)^2 + 9.
\dfrac{3}{2}\ln\left(x^2 - 4x + 20\right) + \dfrac{5}{4}\arctan\dfrac{x - 2}{4} + C. Split \displaystyle 3x - 1 = \frac{3}{2}(2x - 4) + 5; the denominator is (x - 2)^2 + 16.
\dfrac{1}{2}\ln\left(x^2 + 6x + 18\right) - \arctan\dfrac{x + 3}{3} + C. Split \displaystyle x = \frac{1}{2}(2x + 6) - 3; the denominator is (x + 3)^2 + 9, and \displaystyle -3 \cdot \frac{1}{3} = -1.
\dfrac{\pi}{8}. The antiderivative \displaystyle \frac{1}{2}\arctan\frac{x + 1}{2} runs from \displaystyle \frac{1}{2}\arctan 0 = 0 to \displaystyle \frac{1}{2}\arctan 1 = \frac{\pi}{8}.
\displaystyle \dfrac{\pi\sqrt{3}}{9}. The denominator is (x - 2)^2 + 3; the antiderivative \displaystyle \frac{1}{\sqrt{3}}\arctan\frac{x - 2}{\sqrt{3}} gives \displaystyle \frac{1}{\sqrt{3}}\cdot\frac{\pi}{3} - 0, which is \displaystyle \frac{\pi}{3\sqrt{3}}.
\dfrac{\pi}{6}. Here 3 + 2x - x^2 = 4 - (x - 1)^2, and [1, 2] lies inside (-1, 3); \displaystyle \arcsin\frac{x - 1}{2} runs from 0 to \displaystyle \arcsin\frac{1}{2} = \frac{\pi}{6}.
\dfrac{1}{2}\ln 2 + \dfrac{\pi}{8}. Split \displaystyle x + 1 = \frac{1}{2}(2x - 2) + 2; the denominator is (x - 1)^2 + 16, so an antiderivative is \displaystyle \frac{1}{2}\ln\left(x^2 - 2x + 17\right) \displaystyle + \frac{1}{2}\arctan\frac{x - 1}{4}. At 5 it is \displaystyle \frac{1}{2}\ln 32 + \frac{\pi}{8}, at 1 it is \displaystyle \frac{1}{2}\ln 16, and \displaystyle \frac{1}{2}\ln 32 - \frac{1}{2}\ln 16 = \frac{1}{2}\ln 2.
(a) Not an arctangent form: x^2 + 6x + 8 = (x + 3)^2 - 1 = (x + 2)(x + 4) factors, so it needs partial fractions. (b) Not an arctangent form: x^2 + 6x + 9 = (x + 3)^2, and the power rule gives -\dfrac{1}{x + 3} + C on each side of -3. (c) Yes: (x + 3)^2 + 2 gives \displaystyle \dfrac{1}{\sqrt{2}}\arctan\dfrac{x + 3}{\sqrt{2}} + C.
The denominator is (x + 2)^2 + 16, and the derivative of Arez’s function is \displaystyle \frac{1}{4}\cdot\frac{16}{16 + (x + 2)^2}, which is \dfrac{4}{x^2 + 4x + 20}, four times too large. He left out \displaystyle \frac{1}{k}. Correct: \dfrac{1}{4}\arctan\dfrac{x + 2}{4} + C.
The derivative is \displaystyle \dfrac{1}{\sqrt{1 - \frac{(x - 1)^2}{9}}}\cdot\left(-\dfrac{1}{3}\right) = -\dfrac{1}{\sqrt{9 - (x - 1)^2}}, the negative of the integrand. In the derivation, k was moved inside the root as \displaystyle \sqrt{k^2}, which equals \lvert k \rvert; that equals k only for k > 0. With k = -3 the result is off by a sign.
After completing the square the quadratic is (x - h)^2 + k^2 with k > 0, and a square is never negative, so the quadratic is at least k^2 > 0 for every x. A positive quantity is its own absolute value, so \ln\lvert Q \rvert = \ln Q. The same fact means the denominator is never zero, so the integrand and its antiderivative \displaystyle \frac{1}{k}\arctan\frac{x - h}{k} are defined and continuous on the whole line.
\displaystyle \frac{1}{k}\arctan\frac{x - h}{k}\Big|_h^{h + k} \displaystyle = \frac{1}{k}\arctan 1 - \frac{1}{k}\arctan 0 \displaystyle = \frac{\pi}{4k}. The graph of \displaystyle \frac{1}{(x - h)^2 + k^2} is the graph of \displaystyle \frac{1}{x^2 + k^2} slid h units right, and the interval [h, h + k] slides with it from [0, k]. A slide does not change area, which is what Figure 1 shows for k = 1 and h = 2.