Question

Integrate:

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

Answer 1

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

Let $\sqrt{x+1}=t$

$x+1=t^2$

$x=t^2-1$

$\frac{dx}{dt}=\frac{(d{t}^2-1)}{dt}=2t$

$dx=2tdt$

$\int \frac{1}{(t^2-1)t}2t.dt$

$=\int \frac{2dt}{(t^2-1)}$

$=\int \frac{2dt}{(t+1)(t-1)}$

$=\int (\frac{1}{t-1}-\frac{1}{t+1})dt$

$=\int \frac{dt}{t-1}-\int \frac{dt}{t+1}$

$=\left|n\right||t-1|-\left|n\right||t+1|$

$=ln\left|\frac{t-1}{t+1}\right|$

$=ln\left|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right|$ Answers