← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q2a — Step-by-Step Solution

15 marks · Section A

Indefinite integrals · Calculus · asked 7× in 14 yrs · Read the full method →

Question

Evaluate $\displaystyle\int_0^1 \tan^{-1}\!\left(1-\frac1x\right)dx$.

Technique

Integration by parts (with $v=x$); then split the rational integrand into a log-derivative part plus an arctan part via completing the square.

Solution

Let $I=\displaystyle\int_0^1 \tan^{-1}\!\left(1-\frac1x\right)dx=\int_0^1 \tan^{-1}\!\frac{x-1}{x}\,dx$.

Step 1 — Integrate by parts

Take $u=\tan^{-1}\!\left(1-\tfrac1x\right)$, $dv=dx$, so $v=x$ and

Now $\left(1-\tfrac1x\right)^2=\dfrac{(x-1)^2}{x^2}$, so

giving $du=\dfrac{x^2}{2x^2-2x+1}\cdot\dfrac{1}{x^2}\,dx=\dfrac{dx}{2x^2-2x+1}$. Hence

Step 2 — Boundary term vanishes

At $x=1$: $\tan^{-1}(0)=0$, product $=0$. At $x\to0^+$: $1-\tfrac1x\to-\infty$, so $\tan^{-1}\to-\tfrac\pi2$ (finite), and $x\to0$; product $\to0$. Thus the bracket contributes $0$ and

Step 3 — Evaluate $J=\displaystyle\int_0^1\frac{x\,dx}{2x^2-2x+1}$

Split the numerator to expose the derivative of the denominator $(2x^2-2x+1)'=4x-2$:

First integral. $\dfrac14\Big[\ln(2x^2-2x+1)\Big]_0^1=\dfrac14(\ln1-\ln1)=0$.

Second integral. Complete the square: $2x^2-2x+1=2(x-\tfrac12)^2+\tfrac12$, so

Then

So $J=0+\dfrac{\pi}{4}=\dfrac{\pi}{4}$.

Step 4 — Assemble

Answer