#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 40 Maths Textbook Solution.

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

Given:

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

Hint:

Using integration by parts and $\int \frac{1}{1+x^{2}} d x$.

Explanation:

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

$\Rightarrow I=\int \tan ^{-1} t .2 t d t$                                                $\left[\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 \sqrt{x} d t \Rightarrow d x=2 t d t\right]$

$=2 \int \tan ^{-1} t \cdot t d t$

$=2\left[\tan ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right] \ldots\left\{\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\}$

\begin{aligned} &=2 \tan ^{-1} t \cdot \frac{t^{2}}{2}-\int\left(\frac{t^{2}+1-1}{t^{2}+1}\right) d t \\ &=\tan ^{-1} t \cdot t^{2}-\int 1 d t+\int \frac{1}{1+t^{2}} d t \\ &=t^{2} \tan ^{-1} t-t+\tan ^{-1} t+C \\ &=x \tan ^{-1} \sqrt{x}-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\ &=(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C \end{aligned}