Chapter 5 - Review - Exercises - Page 423: 35

$\int \frac{x^3}{1+x^4}~dx = \frac{1}{4}~ln~(1+x^4)+C$

$\int \frac{x^3}{1+x^4}~dx$ Let $u = 1+x^4$ $\frac{du}{dx} = 4x^3$ $dx = \frac{du}{4x^3}$ $\int \frac{x^3}{u}~\frac{du}{4x^3}$ $=\int \frac{1}{4}~\frac{du}{u}$ $= \frac{1}{4}~ln~u+C$ $= \frac{1}{4}~ln~(1+x^4)+C$