TITLE: Solving the integral $\int x\sqrt{1+x}\ dx $
QUESTION [10 upvotes]: I need to solve the following indefinite integral by substitution
$$\int x\sqrt{1+x}\  dx $$with $u=1+x$
Can anyone help me by showing me detailed step by step solution? Thanks in advance!

REPLY [16 votes]: HINT: If $u=1+x$, then $x = u - 1$ and you can substitute those into the integral (of course $du=dx$):
$$\int{(u-1)\sqrt{u}\  du}=\int{(u\sqrt{u}-\sqrt{u})\  du}=\int{(u^{\frac{3}{2}}-u^{\frac{1}{2}})\  du}$$
I won't go further, because I trust that you can finish the last integral on your own. :)

REPLY [6 votes]: Given $u = 1 + x $, then $dx = du $ and so your integral becomes
$$ \int (u-1) \sqrt{u} du = \int u \sqrt{u} du - \int \sqrt{u} du$$
Now, use power rule: 
$$ \int x^n dx = \frac{x^{n+1}}{n+1} + C$$