Integrate $\int \limits _0^1 \tan ^{-1}(x^2+1-x)dx$

$(a)\;\frac{\pi}{2}+1 \\(b)\;\frac{\pi}{2}\\(c)\;\frac{\pi}{2}-1 \\ (d)\;\pi$

$\int \limits_0^1 \large\frac{\pi}{2}$$- \cot ^{-1} (1+x^2-x) => \int\limits_0^1 \large\frac{\pi}{2}$$ - \cot ^{-1} (1+x(-1+x)] dx$
=> $\int\limits_0^1 \large\frac{\pi}{2}$$- \tan ^{-1} \bigg(\large\frac{1}{1+x(1-x)}\bigg)$$dx$
=> $\int\limits_0^1 \large\frac{\pi}{2}$$- \tan ^{-1} \bigg(\large\frac{+x-(-1+x)}{1+x(-1+x)}\bigg)$$dx$
=> $\int \limits _0^1 \large\frac{\pi}{2}$$- \tan ^{-1} [(x-1) - x/1+x(x-1)]dx => \int \limits _0^1 \large\frac{\pi}{2}$$ + \tan ^{-1} (x-1) - \tan ^{-1}(x)dx$
=> $\int \limits _0^1 \large\frac{\pi}{2}$$+ \int \limits_0^1 \tan ^{-1} (x-1) dx -\int \limits_0^1 \tan (x) dx => \int \limits _0^1 \large\frac{\pi}{2}$$dx+ \int \limits_0^1 \tan ^{-1} (x-1) dx -\int \limits_0^1 \tan^{-1} x dx$
$f(x)=f(a-x)$ by this property
=> $\int \limits _0^1 \large\frac{\pi}{2}$$+ \int \limits_0^1 \tan ^{-1} (1-x-1) dx -\int \tan^{-1} (x) dx => \int \limits _0^1 \large\frac{\pi}{2}$$+ 2 \tan ^{-1}(x) dx$
By applying the limits we get,