#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 23

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

Hint:Use substitution method to solve this integral.

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

Solution:

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

\begin{aligned} I &=\int \frac{1}{1+\sqrt{t^{2}}} 2 t d t=\int \frac{2 t}{1+t} d t \\ &=2 \int \frac{t}{1+t} d t=2 \int \frac{1+t-1}{1+t} d t \end{aligned}

\begin{aligned} &=2 \int \frac{(1+t)-1}{1+t} d t=2 \int\left\{\frac{1+t}{1+t}-\frac{1}{1+t}\right\} d t \\ &=2 \int\left\{1-\frac{1}{1+t}\right\} d t=2 \int 1 . d t-2 \int \frac{1}{1+t} d t \end{aligned}

$=2 \int 1 . d t-2 \int \frac{1}{1+t} d t$                            ......$(i)$

$\text { Now } 2 \int 1 . d t=2 \frac{t^{0+1}}{0+1}+c_{1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

$=2 t+c_{1}$                                                          ........$(ii)$

\begin{aligned} &\text { and } 2 \int \frac{1}{1+t} d t \\ &\text { Put } 1+t=p \Rightarrow d t=d p \text { then } \end{aligned}

$2 \int \frac{1}{1+t} d t=2 \int \frac{1}{p} d p=2 \log |p|+c_{2}$

$=2 \log |t+1|+c_{2}$                                        .........$(iii)$

Putting the values of equation (ii) and (iii) in (i) then

\begin{aligned} &I=2 t+c_{1}-\left(2 \log |t+1|+c_{2}\right) \\ &\Rightarrow I=2 t+c_{1}-2 \log |t+1|-c_{2} \\ &\therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c_{1}-c_{2} \end{aligned}

$\therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c \quad\left[\begin{array}{c} \because t^{2}=x \Rightarrow t=\sqrt{x} \\ \text { and } c=c_{1}-c_{2} \end{array}\right]$