## Calculus: Early Transcendentals 8th Edition

$$\int\frac{x}{1+x^4}dx=\frac{\tan^{-1}(x^2)}{2}+C$$
$$A=\int\frac{x}{1+x^4}dx$$ In the substitution exercise with a fraction like this, the first thing you should think is to find out if you can find some substitution strategy to remove the numerator and change the fraction to familiar forms. Here we would do exactly like that: Let $u=x^2$ Then we have $du=2xdx$. That means $xdx=\frac{1}{2}du$ Also, $1+x^4=1+(x^2)^2=1+u^2$ Substitute into $A$: $$A=\frac{1}{2}\int\frac{1}{1+u^2}du$$ $$A=\frac{1}{2}\tan^{-1}u+C$$ $$A=\frac{\tan^{-1}(x^2)}{2}+C$$