## University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{1}{x(x^4+1)}dx=\frac{1}{4}\ln\frac{x^4}{x^4+1}+C$$
$$I=\int\frac{1}{x(x^4+1)}dx$$ Multiply both numerator and denominator with $x^3$: $$I=\int\frac{x^3}{x^4(x^4+1)}dx$$ We set $u=x^4$, so $$du=4x^3dx$$ $$x^3dx=\frac{1}{4}du$$ That means $$I=\frac{1}{4}\int\frac{1}{u(u+1)}du$$ $$I=\frac{1}{4}\int\frac{(u+1)-u}{u(u+1)}du$$ $$I=\frac{1}{4}\Big(\int\frac{du}{u}-\int\frac{du}{u+1}\Big)$$ $$I=\frac{1}{4}(\ln|u|-\ln|u+1|)+C$$ $$I=\frac{1}{4}\ln|\frac{u}{u+1}|+C$$ $$I=\frac{1}{4}\ln|\frac{x^4}{x^4+1}|+C$$ Since $x^4\ge0$ for all $x$, $|\frac{x^4}{x^4+1}|=\frac{x^4}{x^4+1}$ $$I=\frac{1}{4}\ln\frac{x^4}{x^4+1}+C$$