Question

\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]

Answer 1

Let  I  =

\[\int\] (tan^–1 x²) x dx
Putting x² = t
⇒​ 2x dx = dt

\[\Rightarrow \text{ x dx }= \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int 1_{II} . \tan^{- 1_I} t . dt\]
\[ = \frac{1}{2} \tan^{- 1} t\int1 \text{ dt }- \int\left\{ \frac{d}{dt}\left( \tan^{- 1} t \right)\int1 dt \right\}dt\]
\[ = \frac{1}{2} \left[ \tan^{- 1} t . t - \int \frac{t}{1 + t^2}dt \right]\]
\[\text{ Now  putting }\ 1 + t^2 = p\]
\[ \Rightarrow \text{ 2t dt }= dp\]
\[ \Rightarrow \text{ t dt} = \frac{dp}{2}\]
\[ \therefore I = \frac{1}{2}t . \tan^{- 1} t - \frac{1}{2}\int \frac{t dt}{1 + t^2}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{2 x^2} \int \frac{dp}{p}\]
\[ = \frac{t . \tan^{- 1} t}{2} - \frac{1}{4}\ln p + C\]
\[ = \frac{x^2 . \tan^{- 1} x^2}{2} - \frac{1}{4} \text{ ln }\left| 1 + x^4 \right| + C \left[ \because p = 1 + t^2 \right]\]