### Problem asked by Daniel Phulbani

Evaluate $\int x(x^4 + 4x^2 + 4)^{3/2} \;dx$

$\int x(x^4 + 4x^2 + 4)^{3/2} \;dx$

$= \int x((x^2 + 2)^2)^{3/2} \;dx$

$= \int x(x^2 + 2)^3 \;dx$

Substitute $u = x^2 + 2$ and $du = 2x \;dx$ to get

$\int x(x^2 + 2)^3 \;dx$

$= \frac{1}{2} \int (x^2 + 2)^3 2x\;dx$

$= \frac{1}{2} \int u^3 \; du$

$= \frac{1}{2} \frac{u^4}{4} + C$

$= \frac{(x^2 + 1)^4}{8} + C$