Evaluate $\int \sqrt{\frac{a+x}{x}} d x$
to solve this integral we have to apply substitution method for which we are going to put $x=a \cdot \tan ^{2} k$
This means $\mathrm{dx}=2$.a.tank.sec ${ }^{2} k \cdot d k$, then I will be,
$I=\int \sqrt{\frac{\operatorname{asec}^{2} k}{a \tan ^{2} k}} \cdot 2 a \cdot \tan k \cdot \sec ^{2} k \cdot d k=2 a \cdot \operatorname{cosec} k \cdot \tan k \cdot \sec ^{2} k \cdot d k$
In this above integral let $\tan k=t$ then $\sec ^{2} k d k=d t$, put in above equation-
$I=2 a \int \sqrt{\left(t^{2}+1\right)} \cdot d t$
Apply the formula of sqrt $\left(x^{2}+a^{2}\right)=x / 2 \cdot \operatorname{sqrt}\left(a^{2}+x^{2}\right)+a^{2} / 2 \ln
$I=2 a\left[\frac{t}{2} \cdot \sqrt{1+t^{2}}+\frac{1}{2} \cdot \ln \left|t+\sqrt{1+t^{2}}\right|\right]$\left|x+\operatorname{sqrt}\left(a^{2}+x^{2}\right)\right|$
Now put the value of $t$ in above integral $t=\operatorname{tank}$, then finally integral will be-
$I=2 a\left[\frac{\tan k}{2} \cdot \sqrt{1+\tan ^{2} k}+\frac{1}{2} \cdot \ln \mid \tan k+\sqrt{1+\tan ^{2} k}\right]$
Now put the value of $k$ in terms of $x$ that is $\tan ^{2} k=x / a$ in above integral -
$I=2 a\left[\frac{1}{2} \sqrt{\frac{x}{a}} \cdot \sqrt{1+\frac{x}{a}}+\frac{1}{2} \cdot \ln \mid \frac{1}{2} \sqrt{\frac{x}{a}}+\sqrt{1+\frac{x}{a}}\right]$