Answer
$$\int x\sqrt{x+x^{4}}dx=\frac{x^{2}}{2}\sqrt{2+x^{4}}+ln(x^{2}+\sqrt{2+x^{4}})+C$$
Work Step by Step
Using entry #21:
$\int \sqrt{a^{2}+u^{2}}du=\frac{u}{2}\sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2}ln(u+\sqrt{a^{2}+u^{2}})+C$
$$\int x\sqrt{x+x^{4}}dx=\frac{1}{2}\int \sqrt{2+(x^{2})^{2}}d(x^{2})$$
$$=\frac{x^{2}}{2}\sqrt{2+x^{4}}+ln(x^{2}+\sqrt{2+x^{4}})+C$$