Solve this

$u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

We know, $\frac{\mathrm{du}}{\mathrm{dx}}=\frac{2}{1+\mathrm{x}^{2}}$

Using the chain rule of differentiation,

$\frac{d v}{d x}=\frac{1}{1+\left(\frac{2 x}{1+x^{2}}\right)^{2}} \cdot \frac{\left(1+x^{2}\right) \cdot(2 x)^{\prime}-\left(1+x^{2}\right)^{\prime} \cdot(2 x)}{\left(1+x^{2}\right)^{2}}$

$=\frac{\left(1+x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}+(2 x)^{2}} \cdot \frac{2\left(1+x^{2}\right)-(2 x)(2 x)}{\left(1+x^{2}\right)^{2}}$

$=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}+(2 x)^{2}}$

$=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}+(2 x)^{2}}$

Using Chain Rule of Differentiation,

$\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dv}}$

$=\frac{2}{1+x^{2}} \cdot \frac{\left(1+x^{2}\right)^{2}+(2 x)^{2}}{2\left(1-x^{2}\right)}$

$=\frac{\left(1+x^{2}\right)^{2}+(2 x)^{2}}{\left(1+x^{2}\right)\left(1-x^{2}\right)}$

Dividing numerator and denominator by $\left(1+x^{2}\right)^{2}$,

$\frac{d u}{d v}=\frac{1+\left(\frac{2 x}{1+x^{2}}\right)^{2}}{\frac{1-x^{2}}{1+x^{2}}}$

$=\frac{1+\sin ^{2} u}{\cos u}$

$=\sec u(1+\tan u)$ (Ans)