Question

$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$

Answer 1

$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$

$\int \frac{x^4-1}{x^2.x\sqrt{x^2+\frac{1}{x^2}+1}}dx$

$\int \frac{x^4-1}{x^3\sqrt{x^2+\frac{1}{x^2}+1}}dx$

$\int \frac{x}{\sqrt{x^2+\frac{1}{x^2}+1}}dx-\int \frac{1}{x^3\sqrt{x^2+\frac{1}{x^2}+1}}dx$

$=\int \frac{(x-1/x^3)}{\sqrt{x^2+\frac{1}{2}+1}}dx$

Let $x^2+\frac{1}{x^2}+1=t$

$(2x-\frac{1}{x^3})dx=dt$

$\frac{1}{2}\int dt\sqrt{t}=\sqrt{t}+C$

$=\frac{\sqrt{x^4+x^2+1}}{x}+C$