Question

Evaluate: $\int \sqrt{\frac{1+x}{x}}dx$.

Answer 1

We have,

$I=\int \sqrt{\frac{1+x}{x}dx}$

$I=\int \frac{1+x}{\sqrt{x\left(1+x\right)}}dx$

$I=\frac{1}{2}\int \frac{2x+1+1}{\sqrt{x^2+x}}dx$

$I=\frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x}}dx+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x}}$

$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x}}$

$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}+\frac{1}{2}\int \frac{dx}{\sqrt{{\left(x+\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}$

$Put{x}^2+x=t$

$I=\frac{1}{2}\times 2\sqrt{x^2+x}+\frac{1}{2}\ln \left|x+\frac{1}{2}+\sqrt{x^2+x}\right|+c$

$I=\sqrt{x^2+x}+\frac{1}{2}\ln \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x}\right|+c$

Hence, this is the required answer.