Online Test — Integrals
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
$\int\frac{1}{\sqrt{1-x^2}}\,dx=$
$\sin^{-1}x+C$
$\cos^{-1}x+C$
$\tan^{-1}x+C$
$\ln|x|+C$
Explanation: Standard result: $\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C$.
Question 2 of 10
easy
$\int x e^x\,dx=$
$e^x+C$
$(x-1)e^x+C$
$xe^x+C$
$(x+1)e^x+C$
Explanation: IBP: $u=x$, $dv=e^x dx$. $\int xe^x dx=xe^x-e^x+C=(x-1)e^x+C$.
Question 3 of 10
medium
$\int_0^{\pi/2}\frac{\cos x}{1+\sin x}\,dx=$
$\ln 2$
$\pi/4$
$0$
$1$
Explanation: Substitution $t=1+\sin x$: $\int_1^2 dt/t = \ln 2$.
Question 4 of 10
easy
$\int_{-1}^{1}x^3\,dx=$
$0$
$2$
$1/2$
$-1/2$
Explanation: $x^3$ is an odd function. Integral on $[-a,a]=0$.
Question 5 of 10
medium
$\int\frac{1}{x^2-9}\,dx=$
$\frac{1}{6}\ln|\frac{x-3}{x+3}|+C$
$\frac{1}{3}\ln|\frac{x-3}{x+3}|+C$
$\frac{1}{6}\ln|\frac{x+3}{x-3}|+C$
$\tan^{-1}(x/3)+C$
Explanation: $\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$. Here $a=3$: $\frac{1}{6}\ln|\frac{x-3}{x+3}|+C$.
Question 6 of 10
medium
$\int_0^\pi x\sin x\,dx=$
$\pi$
$0$
$2\pi$
$\pi/2$
Explanation: IBP: $[-x\cos x+\sin x]_0^\pi=[-\pi\cos\pi+0]-[0+0]=\pi$.
Question 7 of 10
hard
$\int\frac{x+1}{(x+2)(x+3)}\,dx=$ (using partial fractions)
$-\ln|x+2|+2\ln|x+3|+C$
$\ln|x+2|-\ln|x+3|+C$
$\ln|(x+2)(x+3)|+C$
$\frac{1}{(x+2)(x+3)}+C$
Explanation: $\frac{x+1}{(x+2)(x+3)}=\frac{-1}{x+2}+\frac{2}{x+3}$ (partial fractions). Integrating: $-\ln|x+2|+2\ln|x+3|+C$.
Question 8 of 10
medium
$\int_0^{2a}f(x)dx=0$ if:
$f(-x)=f(x)$
$f(2a-x)=-f(x)$
$f(2a-x)=f(x)$
$f$ is strictly increasing
Explanation: $\int_0^{2a}f(x)dx=0$ when $f(2a-x)=-f(x)$ (the function is 'anti-symmetric' about $x=a$).
Question 9 of 10
medium
$\int_0^{\pi/4}\tan^2 x\,dx=$
$1-\pi/4$
$\pi/4$
$\pi/4-1$
$1+\pi/4$
Explanation: $\tan^2x=\sec^2x-1$. $\int_0^{\pi/4}(\sec^2x-1)dx=[\tan x-x]_0^{\pi/4}=1-\pi/4$.
Question 10 of 10
medium
$\int\ln x\,dx=$
$x\ln x+C$
$x\ln x-x+C$
$\frac{1}{x}+C$
$x\ln x+x+C$
Explanation: IBP: $u=\ln x$, $dv=dx$. $\int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}dx=x\ln x-x+C$.