The integral (x^2)(sqrt{2x+1

[kimmy_koo51]

The integral (x^2)(sqrt[2x+1])dx

Use substitution u= x^2; du= 2xdx Then:

integral u(sqrt[dx+1])

How do you solve from here

 

[galactus]

May I ask what that dx is doing in the radical?. A typo I assume.

\(\displaystyle \int{x^{2}\sqrt{2x+1}}dx\)

Let \(\displaystyle u=\sqrt{2x+1}, \;\ x=\frac{u^{2}-1}{2}, \;\ dx=udu\)

Make the subs and get:

\(\displaystyle \int{u^{2}(\frac{u^{2}-1}{2})^{2}du=\int{[\frac{u^{6}}{4}-\frac{u^{4}}{2}+\frac{u^{2}}{4}]}du\)

Now, it should be easier. Integrate and resub.