Indefinite Integrals — Maths STD 12 Science — Question

Let $\text{x}=\text{t}^2$

$\text{dx}=2\text{t dt}$

$\text{I}=\int2\text{t}\tan^{-1}\text{t dt}$

$=2\Big[\tan^{-1}\text{t}\int\text{t dt}-\int\Big(\frac{1}{1+\text{t}^2}\int\text{t dt}\Big)\text{dt}\Big]$

$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\int\frac{\text{t}^2}{2(1+\text{t}^2)}\text{dt}\Big]$

$=\text{t}^2\tan^{-1}\text{t}-\int\frac{\text{t}^2+1-1}{1+\text{t}^2}\text{dt}$

$=\text{t}^2\tan^{-1}\text{t}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$

$=\text{t}^2\tan^{-1}\text{t}-\text{t}+\tan^{-1}\text{t + C}$

$=(\text{t}^2+1)\tan^{-1}\text{t}-\text{t + C}$

$\text{I}=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$