Solving the integral $\int x\sqrt{1+x}\ dx $

Question

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!

Answer 1

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. :)

Answer 2

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:

$$ \int x^n dx = \frac{x^{n+1}}{n+1} + C$$