Answer
$$\int x(2x^2+3)^4dx=\frac{(2x^2+3)^5}{20}$$
Work Step by Step
To evaluate the integral $\int x(2x^2+3)^4dx$ we will use substitution $u=2x^2+3$ which gives us $du=4xdx\Rightarrow \frac{1}{4}du=xdx.$ Putting this into our integral we get:
$$\int(2x^2+3)^4xdx=\int u^4\cdot\frac{1}{4}du=\frac{1}{4}\int u^4du=\frac{1}{4}\frac{u^5}{5}=\frac{u^5}{20}$$
Now we have to express our solution in terms of $x$ by expressing $u$ in terms of $x$:
$$\int x(2x^2+3)^4dx=\frac{u^5}{20}=\frac{(2x^2+3)^5}{20}$$
