Answer
$$\int\frac{x^3}{(x^4-5)^2}dx=-\frac{1}{4(x^4-5)}+c$$
Work Step by Step
To solve the integral $$\int\frac{x^3}{(x^4-5)^2}dx$$ we will use subtitution $u=x^4-5$ which gives us $du=4x^3dx\Rightarrow \frac{1}{4}du=x^3dx.$ Putting this into the integral we get:
$$\int\frac{x^3}{(x^4-5)^2}dx=\int \frac{1}{u^2}\frac{1}{4}du=\frac{1}{4}\int u^{-2}du=\frac{1}{4}(\frac{u^{-1}}{-1}+c_{1})=-\frac{1}{4}\frac{1}{u}+c$$
where $c$ is arbitrary constant. Now we have to express out solution in terms of $x$ by expressing $u$ in terms of $x$:
$$\int\frac{x^3}{(x^4-5)^2}dx=-\frac{1}{4u}+c=-\frac{1}{4(x^4-5)}+c$$
