Recognise the structure before integrating · source reference

Printed question

Evaluate:

Part	Integral
(a)	$\int\frac{1}{2x}(\ln x)^5\,dx$
(b)	$\int xe^{4-x^2}\,dx$
(c)	$\int(\ln x)^2\,dx$
(d)	$\int\frac4{1+\cos4\theta}\,d\theta$
(e)	$\int\frac1{\sin^2x}e^{\cot x}\,dx$
(f)	$\int\frac1{x^2+2x+5}\,dx$
(g)	$\int\frac{x^2-x}{x^2-16}\,dx$
(h)	$\int\frac{x^2-6x+1}{(3x+1)(x^2+3)}\,dx$
(i)	$\int\frac{e^{2x+1}}{(e^{2x}+1)^3}\,dx$
(j)	$\int\frac{x+1}{\sqrt{1-x^2}}\,dx$

Printed working

(a)

$$\int\frac{(\ln x)^5}{2x}\,dx=\frac12\int\frac1x(\ln x)^5\,dx=\frac12\frac{(\ln x)^6}{6}+c=\frac1{12}(\ln x)^6+c.$$

(b)

$$\int xe^{4-x^2}\,dx=-\frac12\int(-2x)e^{4-x^2}\,dx=-\frac12e^{4-x^2}+c.$$

(c)

$$\begin{aligned}\int(\ln x)^2dx&=x(\ln x)^2-\int x(2\ln x)\frac1x\,dx\\&=x(\ln x)^2-2\int\ln x\,dx\\&=x(\ln x)^2-2\left[x\ln x-\int x\frac1x\,dx\right]\\&=x(\ln x)^2-2x\ln x+2x+c.\end{aligned}$$

(d)

$$\int\frac4{1+\cos4\theta}d\theta=4\int\frac1{1+(2\cos^22\theta-1)}d\theta=2\int\frac1{\cos^22\theta}d\theta=2\int\sec^22\theta\,d\theta=2\frac{\tan2\theta}{2}+c=\tan2\theta+c.$$

(e)

$$\int\frac{e^{\cot x}}{\sin^2x}dx=-\int(-\csc^2x)e^{\cot x}dx=-e^{\cot x}+c.$$

(f)

$$\int\frac{dx}{x^2+2x+5}=\int\frac{dx}{(x+1)^2+2^2}=\frac12\tan^{-1}\frac{x+1}{2}+c.$$

(g)

Long division gives quotient $1$ and remainder $-x+16$ after subtracting $x^2-16$ from $x^2-x$.

$$\begin{aligned}\int\frac{x^2-x}{x^2-16}dx&=\int\left(1+\frac{-x+16}{x^2-16}\right)dx\\&=\int1dx-\frac12\int\frac{2x}{x^2-16}dx+16\int\frac{dx}{x^2-4^2}\\&=x-\frac12\ln|x^2-16|+\frac{16}{2(4)}\ln\left|\frac{x-4}{x+4}\right|+C\\&=x-\frac12\ln|x^2-16|+2\ln\left|\frac{x-4}{x+4}\right|+C.\end{aligned}$$

(h)

$$\frac{x^2-6x+1}{(3x+1)(x^2+3)}=\frac A{3x+1}+\frac{Bx+C}{x^2+3}.$$ $$x^2-6x+1=A(x^2+3)+(Bx+C)(3x+1).$$

Substitute $x=-\frac13$: $\frac19+2+1=A\frac{28}9$, so $A=1$. Substitute $x=0$: $1=3+C$, so $C=-2$. Substitute $x=1$: $1-6+1=4+4(B-2)$, so $B=0$. The printed “Thus” line is:

$$\frac{x^2-6x+1}{(x+1)(x^2+3)}=\frac1{3x+1}-\frac2{x^2+3}.$$

This line drops a 3 in its left denominator. Subsequent integration uses the question’s $(3x+1)(x^2+3)$:

$$\begin{aligned}\int\frac{x^2-6x+1}{(3x+1)(x^2+3)}dx&=\int\left(\frac1{3x+1}-\frac2{x^2+3}\right)dx\\&=\frac13\int\frac3{3x+1}dx-2\int\frac{dx}{x^2+(\sqrt3)^2}\\&=\frac13\ln|3x+1|-\frac2{\sqrt3}\tan^{-1}\frac x{\sqrt3}+c.\end{aligned}$$

(i)

$$\int\frac{e^{2x+1}}{(e^{2x}+1)^3}dx=\int e^{2x}e(e^{2x}+1)^{-3}dx=\frac e2\int2e^{2x}(e^{2x}+1)^{-3}dx=\frac e2\frac{(e^{2x}+1)^{-2}}{-2}+c=-\frac e{4(e^{2x}+1)^2}+c.$$

(j)

$$\begin{aligned}\int\frac{x+1}{\sqrt{1-x^2}}dx&=-\frac12\int(-2x)(1-x^2)^{-1/2}dx+\int\frac{dx}{\sqrt{1-x^2}}\\&=-\frac12\frac{(1-x^2)^{1/2}}{1/2}+\sin^{-1}x+c\\&=-\sqrt{1-x^2}+\sin^{-1}x+c.\end{aligned}$$

Teacher notes

The boxed power formula is $\int f'(x)(f(x))^n dx=(f(x))^{n+1}/(n+1)+c$ ($n\ne-1$). For (a), $f=\ln x$, $f'=1/x$, $n=5$. For (i), $f=e^{2x}+1$, $f'=2e^{2x}$, $n=-3$; for (j), $f=1-x^2$, $f'=-2x$, $n=-1/2$.

The exponential formula is $\int f'(x)e^{f(x)}dx=e^{f(x)}+c$: (b) uses $f=4-x^2$, $f'=-2x$; (e) uses $f=\cot x$, $f'=-\csc^2x$.

Integration by parts: $\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$. For the first integral in (c), $u=(\ln x)^2$, $dv/dx=1$; for the second, $u=\ln x$, $dv/dx=1$.

From MF27: $\cos2A=2\cos^2A-1$, replacing $A$ by $2\theta$ in (d).

Not in MF27: $\int\frac{f'(x)}{a^2+(f(x))^2}dx=\frac1a\tan^{-1}(f(x)/a)+c$; (f) has $f=x+1$, $f'=1$, $a=2$.

(g) uses $\int f'/f=\ln|f|+c$ and MF27 $\int dx/(x^2-a^2)=\frac1{2a}\ln|(x-a)/(x+a)|+c$, where $a=4$. Both logarithms need the modulus sign.

(h): use partial fractions to split a proper fraction into two integrable terms. Then use $\int f'/f=\ln|f|+c$ (modulus sign), and MF27 $\int dx/(x^2+a^2)=a^{-1}\tan^{-1}(x/a)+c$.

(j) also uses MF27 $\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c$, here $a=1$.

Printed answer key

(a) $\frac1{12}(\ln x)^6+c$; (b) $-\frac12e^{4-x^2}+c$; (c) $x(\ln x)^2-2x\ln x+2x+c$; (d) $\tan2\theta+c$; (e) $-e^{\cot x}+c$; (f) $\frac12\tan^{-1}((x+1)/2)+C$; (g) $x-\frac12\ln|x^2-16|+2\ln|(x-4)/(x+4)|+C$; (h) $\frac13\ln|3x+1|-\frac2{\sqrt3}\tan^{-1}(x/\sqrt3)+c$; (i) $-e/[4(e^{2x}+1)^2]+c$; (j) $-\sqrt{1-x^2}+\sin^{-1}x+c$.

Mathematical review

The missing 3 in the “Thus” denominator in (h) is a printed typographical error. The original erroneous line is preserved as text; the decomposition and final answer use 3x+1 and differentiate correctly.