I think this shouldn't be difficult but I can't figure it out! >.<
$\displaystyle \int\displaystyle{\frac{2x}{\sqrt{x+1}}}dx$
Thank you for helping!
Hello !
Substitute $\displaystyle u=\sqrt{x+1}$
$\displaystyle \frac{du}{dx}=\frac{1}{2 \sqrt{x+1}}=\frac 1{2u}$
Thus $\displaystyle dx=2u du$
$\displaystyle \int \frac{2x}{\sqrt{x+1}} ~dx=\int \frac{2x}{u} \times (2u) ~du$
and $\displaystyle x+1=u^2 \implies x=u^2-1$
Finish it ^^
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Otherwise, you can do it easily by parts
$\displaystyle u=x \implies du=dx$
$\displaystyle v'=\frac{dx}{\sqrt{x+1}} \implies v=2 \sqrt{x+1}$