$I=\int_{0}^{1} x \tan ^{-1}\left(\frac{1}{1-x^{2}+x^{4}}\right) d x$
$I=\int_{0}^{1} x \tan ^{-1}\left\{\frac{x^{2}-\left(x^{2}-1\right)}{1+x^{2}\left(x^{2}-1\right)}\right\} d x$
$I=\int_{0}^{1} x\left\{\tan ^{-1} x^{2}-\tan ^{-1}\left(x^{2}-1\right)\right\} d x$
$\text { It } x^{2}=t \Rightarrow 2 x d x=d t$
$\mathrm{I}=\frac{1}{2} \int_{0}^{1}\left\{\tan ^{-1} \mathrm{t}-\tan ^{-1}(\mathrm{t}-1)\right\} \mathrm{dt}$
$=\frac{1}{2} \int_{0}^{1} \tan ^{-1} t d t-\frac{1}{2} \int_{0}^{1} \tan ^{-1}\left(t^{-1}\right) d t$
$=\frac{1}{2} \int_{0}^{1} \tan ^{-1} t d t-\frac{1}{2} \int_{0}^{1} \tan ^{-1}(-t) d t$
$=\frac{1}{2} \int_{0}^{1} \tan ^{-1} t d t+\frac{1}{2} \int_{1}^{1} \tan ^{-1}(t) d t$
$=\int_{0}^{1} \tan ^{-1}(t) d t$
$=\left(t . \tan ^{-1} t\right)_{0}^{1}-\int_{0}^{1} \frac{t}{1+t^{2}} d t$
$=\left(\frac{\pi}{4}\right)-\frac{1}{2}\left[\log \left(1+t^{2}\right)\right]_{0}^{1}$
$=\frac{\pi}{4}-\frac{1}{2} \log _{e}^{2}$
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