Evaluating $\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx$

Question

How to evaluate:

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

Answer 1

HINT:

First put $x^2=y$ in

$$\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx=\int{{x^2 -1}\over{2x^4 \sqrt{2x^4-2x^2+1}}} \mathrm 2xdx$$

to get $$\int \frac{y-1}{2y^2\sqrt{2y^2-2y+1}}dy$$

Now $2y^2-2y+1=\frac{4y^2-4y+2}2=\frac{(2y-1)^2+1}2$

Put $2y-1=\tan\theta$

Answer 2

Let $u=1/x^2$. The integral becomes

$$\frac12 \int du \frac{u-1}{\sqrt{(u-1)^2+1}}$$

I hope you can take it from here. I get as the final result

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

Answer 3

You can write the Integral as

$$ I=\int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} dx $$ Now Put the term inside the Radical as $t$ and Proceed