Answer
$$\int \frac{x^3}{x^4-5} dx=\frac{\ln|x^4-5|}{4}+C$$
Work Step by Step
$$A=\int \frac{x^3}{x^4-5} dx$$
Let $u=x^4-5$
Then $du=(x^4-5)'dx=4x^3dx$. So $x^3dx=\frac{1}{4}du$
Substitute into $A$, we have $$A=\int \frac{1}{u}(\frac{1}{4})du$$ $$A=\frac{1}{4}\int\frac{1}{u}du$$ $$A=\frac{1}{4}\ln|u|+C$$ $$A=\frac{\ln|x^4-5|}{4}+C$$
