\(\int \limits_0^1 \cot^{-1} (1 - x + x^2)dx\)
\(= \int \limits_0^1\tan^{-1} \left(\frac 1{1 - x+ x^2}\right) dx\)
\(= \int \limits_0^1 \tan^{-1} \left(\frac 1{1 - x(1 - x)}\right)dx\)
\(= \int \limits_0^1 \tan^{-1} \left(\frac {(x)+ (1 -x)}{1 - x(1 - x)}\right)dx\) ∵ Property of tan−1(x)
\(= \int \limits_0^1[\tan^{-1}x + \tan^{-1}(1 - x)]dx\)
Taking dummy valuable approach :
Let (1 − x) = t
−dx = dt
x → 0 to 1
⇒ t → 1 to 0
\(= \int \limits_0^1 \tan^{-1}x dx - \int \limits_0^1 \tan^{-1}dt\)
\(= 2\int \limits_0^1 \tan^{-1}x dx\) (Replacing dummy \(\because \int \limits_a^b f(t)dt = \int \limits_b^a b(t)dt\))
\(= 2\left(\frac \pi 4 - \frac{\log e^2}2\right)\)
\(= \frac \pi 4 - \log ^2\)
\(\int\limits_0^1 \tan^{-1}x dx = \int \limits_0^1 \tan^{-1}x.1dx\)
By part integration
\( [x \tan^{-1}x]_0^1 - \int \limits_0^1 x \left(\frac 1{1 + x^2}\right)dx\)
\(= \frac \pi 4 - 0 - \frac 12 \int \limits_0^1 d(\log_e(1 + x^2))\)
\(= \frac \pi 4 - \frac 12[\log_e(1 + x^2)]_0^1\)
\(= \frac \pi 4 - \frac{\log 2}2\)
\(= \frac \pi 2 - \log 2\)
