Question

integeration of 1/x√x^4-1

Answer 1

To find the indefinite integral $\int \frac{1}{x\sqrt{x^4-1}}\, dx$.

Substitute $x^4=t$ then $4x^3dx=dt$

$\int \frac{1}{x\sqrt{x^4-1}}\, dx = \int \frac{4x^3}{4x^4\sqrt{x^4-1}}\, dx \\ \int \frac{1}{x\sqrt{x^4-1}}\, dx = \int \frac{1}{4t\sqrt{t-1}}\, dt$

Now make the substitution $t=\sec^2(\theta),dt=2\sec^2(\theta)\tan(\theta)$

The integral becomes,

$\int \frac{1}{x\sqrt{x^4-1}}\, dx = \int \frac{2\sec^2(\theta)\tan(\theta)}{4\sec^2(\theta)\tan(\theta)}\, d \theta\\ \int \frac{1}{x\sqrt{x^4-1}}\, dx = \int \frac{1}{2}\, d\theta\\ \int \frac{1}{x\sqrt{x^4-1}}\, dx = \frac{\theta}{2}+C\\ \int \frac{1}{x\sqrt{x^4-1}}\, dx = \frac{1}{2}\sec^{-1}(x^2)+C$