Evaluate the following integrals:

Solution:

$\int \tan ^{-1} \sqrt{x} d x$

$\int u . d v=u v-\int v d u$

Choose $u$ in these odder

LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

Here $\mathrm{u}=\tan ^{-1} \sqrt{x}$ and $\mathrm{v}=1$

$\therefore \int \tan ^{-1} \sqrt{x} d x$

$\therefore x \tan ^{-1} \sqrt{x}-\int x \cdot \frac{d\left(\tan ^{-1} \sqrt{x}\right)}{d x}$

$=x \tan ^{-1} \sqrt{x}-\frac{1}{2} \int \frac{\sqrt{x}}{1+x} d x$

Put $\sqrt{x}=t$

$\frac{1}{2 \sqrt{x}} d x=d t$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

and $x=t^{2}$

$=x \tan ^{-1} \sqrt{x}-\int \frac{t^{2}}{1+t^{2}} d t$

$=x \tan ^{-1} \sqrt{x}-\left[\int \frac{1+t^{2}}{1+t^{2}} d t-\int \frac{1}{1+t^{2}} d t\right]$

$=x \tan ^{-1} \sqrt{x}-\left[\sqrt{x}-\tan ^{-1} \sqrt{x}\right]+c$