#### provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 70

Answer: $\frac{1}{2} \log |1+\sqrt{x}|+c$

Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{\sqrt{x}+x} d x$

Solution:

\begin{aligned} &\text { Let } I=\int \frac{1}{\sqrt{x}+x} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ &\Rightarrow d x=2 \sqrt{x} \; d t \text { then } \end{aligned}

\begin{aligned} &I=\int \frac{1}{t+t^{2}} 2 \sqrt{x} \; d t \\ &=\int \frac{1}{t(1+t)} 2 \sqrt{x} \; d t=2 \int \frac{1}{1+t} d t \end{aligned}                $\left[\begin{array}{l} \because \sqrt{x}=t \\ \Rightarrow x=t^{2} \end{array}\right]$

\begin{aligned} &\text { Again put } 1+t=u \Rightarrow d t=d u \text { then }\\ &I=2 \int \frac{1}{u} d u=\frac{1}{2} \log |u|+c \end{aligned}$\left[\because \int \frac{1}{x} d x=\log |x|+c\right]$

$=\frac{1}{2} \log |t+1|+c \quad[\because u=1+t]$

$=\frac{1}{2} \log |1+\sqrt{x}|+c \quad[\because t=\sqrt{x}]$