Answer
$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}(x^3+1)^{3/2}$$
Work Step by Step
To find the integral $\int x^2\sqrt{x^3+1}dx$ we will use substitution $u=x^3+1$ which gives us $du=3x^2dx\Rightarrow \frac{1}{3}du=x^2dx.$ Putting this into our integral we get:
$$\int x^2\sqrt{x^3+1}dx=\int\sqrt u\cdot\frac{1}{3}du=\frac{1}{3}\int u^{1/2}du=\frac{1}{3}\frac{u^{3/2}}{\frac{3}{2}}=\frac{1}{3}\frac{2}{3}u^{3/2}=\frac{2}{9}u^{3/2}$$
Now we have to express solution in terms of $x$ by expressing $u$ in terms of $x$:
$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}u^{3/2}=\frac{2}{9}(x^3+1)^{3/2}$$
