#### Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 66

Answer: $2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+c$

Hint: Use substitution method to solve this integral

Given: $\int \sqrt{e^{x}-1} \; d x$

Solution:

\begin{aligned} &\operatorname{let} I=\int \sqrt{e^{x}-1} d x \\ &\text { Putting } \mathrm{e}^{x}-1=t^{2} \Rightarrow e^{x} d x=2 t d t \Rightarrow d x=\frac{2 t \cdot d t}{e^{x}} \text { then } \end{aligned}

$I=\int \sqrt{t^{2}} \frac{2 t \; d t}{e^{x}}$

$=2 \int \frac{t \cdot t}{t^{2}+1} d t \quad\left[\because e^{x}-1=t^{2} \Rightarrow t^{2}+1=e^{x}\right]$

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

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

$=2 \int 1 d t-2 \int \frac{1}{t^{2}+1} d t=2 \int t^{0} d t-2 \int \frac{1}{t^{2}+1} d t$

$=2 \frac{t^{0+1}}{0+1}-2 \tan ^{-1}(t)+c$                        $\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c \end{array}\right]$

\begin{aligned} &=2 t-2 \tan ^{-1}(t)+c \\ &=2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+\mathrm{c} \end{aligned}        $\left[\because t^{2}=e^{x}-1 \Rightarrow t=\sqrt{e^{x}-1}\right]$