Browse Questions

# Choose the correct answers in the value of $\int\limits_0^1\tan^{-1}\bigg(\frac{\large 2x-1}{\large 1+x-x^2}\bigg)dx\;is$$(A)\;1\qquad(B)\;0\qquad(C)\;-1\qquad(D)\;\frac{\pi}{4}$

$(a)\;1\qquad(b)\;0\qquad(c)\;-1\qquad(d)\;\frac{\pi}{4}$

Toolbox:
• (i) $\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii) $\int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
Given $I=\int \limits_0^1 \tan ^{-1}\bigg(\frac{2x-1}{1+x-x^2}\bigg) dx$

2x-1 can be written as $x+x-1$ and

$1+x-x^2$ can be written as $1+x(1-x)$

Therefore $I=\int \limits_0^1 \tan ^{-1} \bigg(\frac{x-(1-x)}{1+x(1-x)}\bigg) dx$

This is of the form $\tan ^{-1}\bigg[\frac{x-y}{1+xy}\bigg]$

Therefore $I=\int \limits_0^1 \tan ^{-1}x-\tan ^{-1}(1-x) dx$ ------(1)

Now applying the property $\int \limits_a^b f(x)dx=\int \limits_a^b f(a-x)dx$

$I=\int \limits_0^1 \tan ^{-1} (1-x)-\tan ^{-1} (1-1+x)dx$

$=\int \limits_0^1 \tan (1-x) -\tan^{-1}(x) dx$ ------(2)

$2I=\int \limits_0^1 [\tan ^{-1}x -\tan ^{-1}(1-x)+\tan (1-x)-\tan ^{-1}x]dx$

$I=0$

Hence $I=0$