Special Forms, Partial Fractions and Integration by Parts — Worked Solutions

Ques­tion 1: Three direct spe­cial forms#

The prob­lem#

Find (a) $\displaystyle \int\frac{dx}{x^2 - 25}$; (b) $\displaystyle \int\frac{dx}{4 + 9x^2}$; (c) $\displaystyle \int\frac{dx}{\sqrt{x^2 + 16}}$.

Under­stand­ing the prob­lem#

Each inte­grand matches one of the stan­dard forms in the les­son once you iden­tify the con­stant $a$. In (b) the coef­fi­cient $9$ on $x^2$ must be taken out first.

The idea#

Use

$$\displaystyle \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left\lvert\frac{x - a}{x + a}\right\rvert,\quad \int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa,\quad \int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert.$$

Step-by-step solu­tion#

Part (a)

Step 1. $x^2 - 25 = x^2 - 5^2$, so $a = 5$.

Step 2. Apply the for­mula.

$$\displaystyle \int\frac{dx}{x^2 - 25} = \frac{1}{2 \times 5}\log\left\lvert\frac{x - 5}{x + 5}\right\rvert + C = \frac{1}{10}\log\left\lvert\frac{x - 5}{x + 5}\right\rvert + C.$$

Part (b)

Step 1. Take out the $9$ so that $x^2$ has coef­fi­cient $1$.

$$\displaystyle \int\frac{dx}{4 + 9x^2} = \frac19\int\frac{dx}{x^2 + \frac49} = \frac19\int\frac{dx}{x^2 + \left(\frac23\right)^2}.$$

Step 2. Now $\displaystyle a = \frac23$, and $\displaystyle \frac1a = \frac32$.

$$\displaystyle \frac19 \cdot \frac32\tan^{-1}\frac{x}{2/3} + C = \frac16\tan^{-1}\frac{3x}{2} + C.$$

Part (c)

Step 1. $x^2 + 16 = x^2 + 4^2$, so $a = 4$.

Step 2. Apply the for­mula.

$$\displaystyle \int\frac{dx}{\sqrt{x^2 + 16}} = \log\left\lvert x + \sqrt{x^2 + 16}\right\rvert + C.$$

Check­ing the answer#

(b) Dif­fer­en­ti­ate: $\displaystyle \frac16 \cdot \frac{1}{1 + \frac{9x^2}{4}} \cdot \frac32 = \frac14 \cdot \frac{4}{4 + 9x^2} = \frac{1}{4 + 9x^2}$. Cor­rect.

Answer#

(a) $\displaystyle \frac{1}{10}\log\left\lvert\frac{x - 5}{x + 5}\right\rvert + C$; (b) $\displaystyle \frac16\tan^{-1}\frac{3x}{2} + C$; (c) $\log\left\lvert x + \sqrt{x^2 + 16}\right\rvert + C$.

Com­mon mis­take to avoid#

In (b), do not use $a = 2$ and for­get the $9$. The for­mula needs the form $x^2 + a^2$ with coef­fi­cient $1$ on $x^2$.

Ques­tion 2: Com­plet­ing the square#

The prob­lem#

Find (a) $\displaystyle \int\frac{dx}{x^2 - 4x + 13}$; (b) $\displaystyle \int\frac{dx}{\sqrt{5 - 4x - x^2}}$.

Under­stand­ing the prob­lem#

The qua­drat­ics are not yet in a stan­dard form. After com­plet­ing the square each becomes $(x - h)^2 + a^2$ or $a^2 - (x - h)^2$.

The idea#

Com­plete the square, then use the $\tan^{-1}$ form for (a) and the $\sin^{-1}$ form for (b), with $x - h$ in place of $x$.

Step-by-step solu­tion#

Part (a)

Step 1. Com­plete the square: half of $-4$ is $-2$.

$$x^2 - 4x + 13 = (x^2 - 4x + 4) + 9 = (x - 2)^2 + 3^2.$$

Step 2. Apply $\displaystyle \int\frac{dx}{X^2 + a^2} = \frac1a\tan^{-1}\frac Xa$ with $X = x - 2$, $a = 3$.

$$\displaystyle \int\frac{dx}{(x - 2)^2 + 3^2} = \frac13\tan^{-1}\frac{x - 2}{3} + C.$$

Part (b)

Step 1. Take out a minus sign from the $x$ terms and com­plete the square.

$$\begin{aligned} 5 - 4x - x^2 &= 5 - (x^2 + 4x) \\ &= 5 - \bigl[(x + 2)^2 - 4\bigr] \\ &= 9 - (x + 2)^2 = 3^2 - (x + 2)^2. \end{aligned}$$

Step 2. Apply $\displaystyle \int\frac{dx}{\sqrt{a^2 - X^2}} = \sin^{-1}\frac Xa$ with $X = x + 2$, $a = 3$.

$$\displaystyle \int\frac{dx}{\sqrt{9 - (x + 2)^2}} = \sin^{-1}\frac{x + 2}{3} + C.$$

Check­ing the answer#

(b) Dif­fer­en­ti­ate: $\displaystyle \frac{1}{\sqrt{1 - \frac{(x + 2)^2}{9}}} \cdot \frac13 = \frac{1}{\sqrt{9 - (x + 2)^2}}$, and $9 - (x + 2)^2 = 5 - 4x - x^2$. Cor­rect.

Answer#

(a) $\displaystyle \frac13\tan^{-1}\frac{x - 2}{3} + C$; (b) $\displaystyle \sin^{-1}\frac{x + 2}{3} + C$.

Ques­tion 3: A lin­ear term over a qua­dratic#

The prob­lem#

Find $\displaystyle \int\frac{3x - 2}{x^2 + 2x + 5}\,dx$.

Under­stand­ing the prob­lem#

The numer­a­tor is lin­ear and the denom­i­na­tor a qua­dratic that does not fac­torise ($2^2 - 20 < 0$). This is the pat­tern of Exam­ple 4 in the les­son.

The idea#

Write the numer­a­tor as (a mul­ti­ple of the deriv­a­tive of the denom­i­na­tor) $+$ (a con­stant). The first part inte­grates to a $\log$; the sec­ond, after com­plet­ing the square, to a $\tan^{-1}$.

Step-by-step solu­tion#

Step 1. The deriv­a­tive of the denom­i­na­tor is $2x + 2$. Find $\lambda$ and $\mu$ with $3x - 2 = \lambda(2x + 2) + \mu$.

Com­par­ing coef­fi­cients of $x$: $\displaystyle 2\lambda = 3 \Rightarrow \lambda = \frac32$. Con­stants: $2\lambda + \mu = -2 \Rightarrow 3 + \mu = -2 \Rightarrow \mu = -5$.

$$\displaystyle 3x - 2 = \frac32(2x + 2) - 5.$$

Step 2. Split the inte­gral.

$$\displaystyle \int\frac{3x - 2}{x^2 + 2x + 5}\,dx = \frac32\int\frac{2x + 2}{x^2 + 2x + 5}\,dx - 5\int\frac{dx}{x^2 + 2x + 5}.$$

Step 3. First inte­gral: the numer­a­tor is the deriv­a­tive of the denom­i­na­tor, so it gives a log­a­rithm (the denom­i­na­tor is always pos­i­tive, so no mod­u­lus is needed).

$$\displaystyle \frac32\log(x^2 + 2x + 5).$$

Step 4. Sec­ond inte­gral: com­plete the square, $x^2 + 2x + 5 = (x + 1)^2 + 2^2$.

$$\displaystyle 5\int\frac{dx}{(x + 1)^2 + 2^2} = \frac52\tan^{-1}\frac{x + 1}{2}.$$

Step 5. Com­bine.

$$\displaystyle \int\frac{3x - 2}{x^2 + 2x + 5}\,dx = \frac32\log(x^2 + 2x + 5) - \frac52\tan^{-1}\frac{x + 1}{2} + C.$$

Check­ing the answer#

Dif­fer­en­ti­ate: $\displaystyle \frac32 \cdot \frac{2x + 2}{x^2 + 2x + 5} - \frac52 \cdot \frac{2}{(x + 1)^2 + 4} = \frac{3x + 3 - 5}{x^2 + 2x + 5} = \frac{3x - 2}{x^2 + 2x + 5}$. Cor­rect.

Answer#

$\displaystyle \frac32\log(x^2 + 2x + 5) - \frac52\tan^{-1}\frac{x + 1}{2} + C$.

Ques­tion 4: Par­tial frac­tions#

The prob­lem#

Find (a) $\displaystyle \int\frac{x + 4}{x^2 - 3x + 2}\,dx$; (b) $\displaystyle \int\frac{2x}{(x + 1)(x^2 + 1)}\,dx$.

Under­stand­ing the prob­lem#

Both are proper ratio­nal func­tions (the top has lower degree than the bot­tom). In (a) the denom­i­na­tor fac­torises into two lin­ear fac­tors; in (b) there is one lin­ear fac­tor and one qua­dratic fac­tor that does not fac­torise.

The idea#

Split into par­tial frac­tions using the pat­terns in the les­son, find the con­stants, and inte­grate each piece.

Step-by-step solu­tion#

Part (a)

Step 1. Fac­torise: $x^2 - 3x + 2 = (x - 1)(x - 2)$. Write

$$\displaystyle \frac{x + 4}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \;\Rightarrow\; x + 4 = A(x - 2) + B(x - 1).$$

Step 2. Put $x = 1$: $5 = A(-1)$, so $A = -5$. Put $x = 2$: $6 = B(1)$, so $B = 6$.

Step 3. Inte­grate.

$$\displaystyle \int\left(\frac{-5}{x - 1} + \frac{6}{x - 2}\right)dx = -5\log\lvert x - 1 \rvert + 6\log\lvert x - 2 \rvert + C.$$

Part (b)

Step 1. Write

$$\displaystyle \frac{2x}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \;\Rightarrow\; 2x = A(x^2 + 1) + (Bx + C)(x + 1).$$

Step 2. Put $x = -1$: $-2 = A(2)$, so $A = -1$.

Step 3. Com­pare coef­fi­cients. $x^2$: $0 = A + B$, so $B = 1$. Con­stant term: $0 = A + C$, so $C = 1$.

$$\displaystyle \frac{2x}{(x + 1)(x^2 + 1)} = \frac{-1}{x + 1} + \frac{x + 1}{x^2 + 1}.$$

Step 4. Split the sec­ond frac­tion as $\displaystyle \frac{x}{x^2 + 1} + \frac{1}{x^2 + 1}$ and inte­grate each part.

$$\displaystyle \begin{aligned} \int\frac{-1}{x + 1}\,dx &= -\log\lvert x + 1 \rvert, \\ \int\frac{x}{x^2 + 1}\,dx &= \frac12\log(x^2 + 1), \\ \int\frac{1}{x^2 + 1}\,dx &= \tan^{-1}x. \end{aligned}$$

Step 5. Com­bine.

$$\displaystyle -\log\lvert x + 1 \rvert + \frac12\log(x^2 + 1) + \tan^{-1}x + C.$$

Check­ing the answer#

(b) Check the $x$ coef­fi­cient too: $B + C = 1 + 1 = 2$, which matches $2x$ on the left. All three coef­fi­cients agree.

Answer#

(a) $6\log\lvert x - 2 \rvert - 5\log\lvert x - 1 \rvert + C$; (b) $\displaystyle -\log\lvert x + 1 \rvert + \frac12\log(x^2 + 1) + \tan^{-1}x + C$.

Ques­tion 5: Inte­gra­tion by parts#

The prob­lem#

Find (a) $\displaystyle \int x\sin 3x\,dx$; (b) $\displaystyle \int x\log(2x)\,dx$; (c) $\displaystyle \int x^2e^{-x}\,dx$.

Under­stand­ing the prob­lem#

Each inte­grand is a prod­uct of two dif­fer­ent kinds of func­tion, so inte­gra­tion by parts is needed.

The idea#

$$\displaystyle \int u\,v\,dx = u\int v\,dx - \int\left(u'\int v\,dx\right)dx,$$

choos­ing $u$ by ILATE (Inverse trig, Log, Alge­braic, Trig, Expo­nen­tial). In (a) and (c), $u$ is the alge­braic fac­tor; in (b), $u = \log(2x)$.

Step-by-step solu­tion#

Part (a)

Step 1. Take $u = x$, $v = \sin 3x$. Then $u' = 1$ and $\displaystyle \int\sin 3x\,dx = -\frac{\cos 3x}{3}$.

Step 2. Apply the for­mula.

$$\displaystyle \begin{aligned} \int x\sin 3x\,dx &= x\left(-\frac{\cos 3x}{3}\right) - \int 1 \cdot \left(-\frac{\cos 3x}{3}\right)dx \\ &= -\frac{x\cos 3x}{3} + \frac13\int\cos 3x\,dx \\ &= -\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C. \end{aligned}$$

Part (b)

Step 1. Take $u = \log(2x)$ (L comes before A), $v = x$. Then $\displaystyle u' = \frac{2}{2x} = \frac1x$ and $\displaystyle \int x\,dx = \frac{x^2}{2}$.

Step 2. Apply the for­mula.

$$\displaystyle \begin{aligned} \int x\log(2x)\,dx &= \frac{x^2}{2}\log(2x) - \int\frac1x \cdot \frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\log(2x) - \int\frac x2\,dx \\ &= \frac{x^2}{2}\log(2x) - \frac{x^2}{4} + C. \end{aligned}$$

Part (c)

Step 1. Take $u = x^2$, $v = e^{-x}$. Then $u' = 2x$ and $\displaystyle \int e^{-x}\,dx = -e^{-x}$.

$$\displaystyle \int x^2e^{-x}\,dx = -x^2e^{-x} + \int 2xe^{-x}\,dx.$$

Step 2. The new inte­gral still has a prod­uct, so use parts again with $u = 2x$, $v = e^{-x}$.

$$\displaystyle \int 2xe^{-x}\,dx = -2xe^{-x} + \int 2e^{-x}\,dx = -2xe^{-x} - 2e^{-x}.$$

Step 3. Com­bine and fac­tor.

$$\displaystyle \int x^2e^{-x}\,dx = -x^2e^{-x} - 2xe^{-x} - 2e^{-x} + C = -e^{-x}(x^2 + 2x + 2) + C.$$

Check­ing the answer#

(c) Dif­fer­en­ti­ate: $e^{-x}(x^2 + 2x + 2) - e^{-x}(2x + 2) = x^2e^{-x}$. Cor­rect.

Answer#

(a) $\displaystyle -\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C$; (b) $\displaystyle \frac{x^2}{2}\log(2x) - \frac{x^2}{4} + C$; (c) $-e^{-x}(x^2 + 2x + 2) + C$.

Com­mon mis­take to avoid#

In (b), choos­ing $u = x$ would force you to inte­grate $\log(2x)$ first, which is harder. Fol­low ILATE.

Ques­tion 6: $\sin^{-1}x$ and an $e^x(f + f')$ inte­gral#

The prob­lem#

Find (a) $\displaystyle \int\sin^{-1}x\,dx$; (b) $\displaystyle \int e^x\left(\tan x + \sec^2 x\right)dx$.

Under­stand­ing the prob­lem#

(a) has only one func­tion, but you can treat it as $\sin^{-1}x \times 1$ and use parts. (b) has the spe­cial pat­tern $e^x(f + f')$.

The idea#

(a) Parts with $u = \sin^{-1}x$, $v = 1$. (b) Use $\displaystyle \int e^x\bigl(f(x) + f'(x)\bigr)dx = e^xf(x) + C$ with $f(x) = \tan x$.

Step-by-step solu­tion#

Part (a)

Step 1. Take $u = \sin^{-1}x$, $v = 1$. Then $\displaystyle u' = \frac{1}{\sqrt{1 - x^2}}$ and $\displaystyle \int 1\,dx = x$.

$$\displaystyle \int\sin^{-1}x\,dx = x\sin^{-1}x - \int\frac{x}{\sqrt{1 - x^2}}\,dx.$$

Step 2. For the remain­ing inte­gral put $t = 1 - x^2$, so $dt = -2x\,dx$.

$$\displaystyle \int\frac{x}{\sqrt{1 - x^2}}\,dx = -\frac12\int t^{-1/2}\,dt = -\sqrt t = -\sqrt{1 - x^2}.$$

Step 3. Sub­sti­tute back.

$$\displaystyle \int\sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1 - x^2} + C.$$

Part (b)

Step 1. Let $f(x) = \tan x$. Then $f'(x) = \sec^2 x$, so the inte­grand is exactly $e^x(f + f')$.

Step 2. Apply the rule.

$$\displaystyle \int e^x(\tan x + \sec^2 x)\,dx = e^x\tan x + C.$$

Check­ing the answer#

(a) Dif­fer­en­ti­ate: $\displaystyle \sin^{-1}x + \frac{x}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \sin^{-1}x$. (b) Dif­fer­en­ti­ate: $e^x\tan x + e^x\sec^2 x$. Both cor­rect.

Answer#

(a) $x\sin^{-1}x + \sqrt{1 - x^2} + C$; (b) $e^x\tan x + C$.

Ques­tion 7: $\sqrt{16 - x^2}$#

The prob­lem#

Find $\displaystyle \int\sqrt{16 - x^2}\,dx$.

Under­stand­ing the prob­lem#

This is the form $\sqrt{a^2 - x^2}$ with $a = 4$.

The idea#

Use the lesson's for­mula

$$\displaystyle \int\sqrt{a^2 - x^2}\,dx = \frac x2\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac xa + C.$$

Step-by-step solu­tion#

Step 1. Iden­tify $a = 4$, so $a^2 = 16$ and $\displaystyle \frac{a^2}{2} = 8$.

Step 2. Sub­sti­tute.

$$\displaystyle \int\sqrt{16 - x^2}\,dx = \frac x2\sqrt{16 - x^2} + 8\sin^{-1}\frac x4 + C.$$

Check­ing the answer#

Over $[-4, 4]$ this should give the area of a half-disc of radius $4$, which is $8\pi$. The for­mula gives $\displaystyle 8 \cdot \frac{\pi}{2} - 8 \cdot \left(-\frac{\pi}{2}\right) = 8\pi$. Cor­rect.

Answer#

$\displaystyle \frac x2\sqrt{16 - x^2} + 8\sin^{-1}\frac x4 + C$.

Ques­tion 8: $e^{2x}\sin x$ by parts twice#

The prob­lem#

Find $\displaystyle \int e^{2x}\sin x\,dx$ (use parts twice).

Under­stand­ing the prob­lem#

Nei­ther fac­tor dis­ap­pears on dif­fer­en­ti­a­tion: $\sin x$ turns into $\cos x$ and back, $e^{2x}$ stays expo­nen­tial. After two appli­ca­tions of parts the orig­i­nal inte­gral reap­pears, and you solve for it like an unknown.

The idea#

Call the inte­gral $I$. Apply parts twice, always keep­ing the trigono­met­ric fac­tor as $u$ (T before E in ILATE). You will get an equa­tion of the form $I = (\text{known}) - kI$.

Step-by-step solu­tion#

Step 1. Let $\displaystyle I = \int e^{2x}\sin x\,dx$. Take $u = \sin x$, $v = e^{2x}$, so $\displaystyle \int e^{2x}\,dx = \frac{e^{2x}}{2}$.

$$\displaystyle I = \frac{e^{2x}}{2}\sin x - \frac12\int e^{2x}\cos x\,dx.$$

Step 2. Apply parts to $\displaystyle \int e^{2x}\cos x\,dx$ with $u = \cos x$, $u' = -\sin x$.

$$\displaystyle \int e^{2x}\cos x\,dx = \frac{e^{2x}}{2}\cos x + \frac12\int e^{2x}\sin x\,dx = \frac{e^{2x}}{2}\cos x + \frac12 I.$$

Step 3. Sub­sti­tute into Step 1.

$$\displaystyle I = \frac{e^{2x}}{2}\sin x - \frac{e^{2x}}{4}\cos x - \frac14 I.$$

Step 4. Move $\displaystyle \frac14 I$ to the left and solve.

$$\displaystyle \begin{aligned} \frac54 I &= \frac{e^{2x}}{4}(2\sin x - \cos x) \\ I &= \frac{e^{2x}}{5}(2\sin x - \cos x) + C. \end{aligned}$$

Check­ing the answer#

Dif­fer­en­ti­ate: $\displaystyle \frac{2e^{2x}}{5}(2\sin x - \cos x) + \frac{e^{2x}}{5}(2\cos x + \sin x) = \frac{e^{2x}}{5}(5\sin x) = e^{2x}\sin x$. Cor­rect.

Answer#

$\displaystyle \int e^{2x}\sin x\,dx = \frac{e^{2x}}{5}(2\sin x - \cos x) + C$.

Com­mon mis­take to avoid#

In the sec­ond appli­ca­tion, keep the same choice ($u$ = the trigono­met­ric fac­tor). Switch­ing to $u = e^{2x}$ undoes the first step and gives $I = I$.