# Thread: Find the indefinite integral

1. ## Find the indefinite integral

Hi again

$\int \frac{\sqrt{x}}{x-1}~dx$

it seems none of the u subsitution of $\sqrt{x}, x-1, x$ works

some hints would be greatly appreciated =)

2. Originally Posted by ugkwan
Hi again

$\int \frac{\sqrt{x}}{x-1}~dx$

it seems none of the u subsitution of $\sqrt{x}, x-1, x$ works

some hints would be greatly appreciated =)
$x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)$

$\displaystyle\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}=\frac{2\sqrt{x}}{x-1}$

Hence you can convert your integral to an integral of the sum of partial fractions

3. Originally Posted by ugkwan
Hi again

$\int \frac{\sqrt{x}}{x-1}~dx$

it seems none of the u subsitution of $\sqrt{x}, x-1, x$ works

some hints would be greatly appreciated =)
Try $u=\sqrt{x}$ to get:

$\displaystyle \int \dfrac{2u^2}{u^2-1} \; du$

Now try integration by parts splitting the integrand into $2u$ and $u/(u^2-1)$

CB

4. Originally Posted by ugkwan
Hi again

$\int \frac{\sqrt{x}}{x-1}~dx$

it seems none of the u subsitution of $\sqrt{x}, x-1, x$ works

some hints would be greatly appreciated =)
In terms of the substitution, an alternative is

$u=\sqrt{x}$

$u^2=x\Rightarrow\frac{d}{dx}u^2=\frac{d}{dx}x\Righ tarrow\frac{d}{du}u^2\frac{du}{dx}=1\Rightarrow\ 2u\frac{du}{dx}=1$

$dx=2udu$

$\displaystyle\int{\frac{\sqrt{x}}{x^2-1}dx=\int{\frac{u}{u^2-1}2u}du=\int{\frac{2u^2}{u^2-1}}du$

$=\displaystyle\int{\frac{2u^2}{(u+1)(u-1)}}du$

$\displaystyle\frac{u}{u-1}+\frac{u}{u+1}=\frac{2u^2}{u^2-1}$

$u-1=y\Rightarrow\ du=dy,\;\;\;u=y+1$

$w=u+1\Rightarrow\ dw=du,\;\;\;u=w-1$

leaving us with

$\displaystyle\int{\frac{y+1}{y}}dy+\int{\frac{w-1}{w}}dw$