How Can I resolve this Integral?

Question

I do not know if apply parts, or do a change:

http://i.snag.gy/eeZam.jpg

$\int x(2x+1)^{7/2}\, dx$

Answer 1

Using integration by parts with $u=x$ and $v=\frac19 (2x+1)^{9/2}$, we have

$$\begin{align} \int x(2x+1)^{7/2}\,dx&=\frac19 x(2x+1)^{9/2}-\frac19\int (2x+1)^{9/2}\,dx\\\\ &=\frac19 x(2x+1)^{9/2}-\frac1{99}(2x+1)^{11/2}+C \end{align}$$

Answer 2

If you make the change of variable $$ u=2x+1 $$ then you get $$ \int x(2x+1)^{7/2}dx=\frac14\int \left(u-1\right)u^{7/2}du=\frac14\int u^{9/2}du-\frac14\int u^{7/2}du. $$

Answer 3

This can be done with substitution.

$$u=(2x+1)$$ $$du=2dx$$

$\int x(2x+1)^\frac{7}{2}dx=\frac{1}{2}\int\frac{1}{2}(u-1)u^\frac{7}{2}du=\frac{1}{4}\int u^\frac{9}{2}-u^\frac{7}{2}du=\frac{1}{4}(\frac{2}{11}u^\frac{11}{2}-\frac{2}{9}u^\frac{9}{2})+c$ $$$$ I'll leave the rest for you to substitute and simplify.

Answer 4

Use a simple substitution. Let $u = 2x + 7.$ Then we have $du = 2dx.$ Our integral then becomes $\int \frac{1}{2}u^{\frac{7}{2}} du.$ We know this to be $\frac{1}{5}u^{\frac{5}{2}} + C.$ Substituting back in, we have our answer, $\boxed{\frac{1}{5}(2x + 1)^{\frac{5}{2}} + C}.$