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

$$\int \frac{dx}{x\sqrt{x^4-1}}=\frac{1}{2}\tan^{-1}\sqrt{x^4-1}+C$$ Or: $$\frac{1}{2}\sec^{-1}(x^2)+C$$
$$A=\int \frac{dx}{x\sqrt{x^4-1}}=\int\frac{x^3}{x^4\sqrt{x^4-1}}dx$$ We set $u=\sqrt{x^4-1}$. That makes $u^2=x^4-1$, and thus, $x^4=u^2+1$ Then $$du=\frac{(x^4-1)'}{2\sqrt{x^4-1}}dx=\frac{4x^3}{2\sqrt{x^4-1}}dx=\frac{2x^3}{\sqrt{x^4-1}}dx$$ $$\frac{x^3}{\sqrt{x^4-1}}dx=\frac{1}{2}du$$ Therefore, $$A=\frac{1}{2}\int\frac{1}{u^2+1}du$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$ So $$A=\frac{1}{2}\times\frac{1}{1}\tan^{-1}\Big(\frac{u}{1}\Big)+C$$ $$A=\frac{1}{2}\tan^{-1}u+C$$ $$A=\frac{1}{2}\tan^{-1}\sqrt{x^4-1}+C$$ Or alternatively (using a different integration strategy), we can arrive at: $$\frac{1}{2}\sec^{-1}(x^2)+C$$