Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
 A: Substitute $t=\frac1{x^2}$ to get
$$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} {dx}
 =\frac12 \int \frac{(t-1)dt}{\sqrt{2-2t+t^2}}
=\frac12 \sqrt{2-2t+t^2}+C
$$
A: Following @AdityaDwivedi's suggestion, $dt=4x^{-5}(x^2-1)dx$ so the integral is $\int\frac{dt}{4\sqrt{2+t}}=\tfrac12\sqrt{2+t}+C$. As to how you'd come up with this idea, note the original integral is $\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$, which suggests the proposed substitution, or better still $u=2-2x^{-2}+x^{-4}$.
A: $$I = \int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx$$
Take $x^4$ factor out from the square root.
$$ = \int\frac{x^2-1}{x^5\sqrt{2-2x^{-2}+x^{-4}}}$$
Now substitute $\sqrt{2-2x^2+x^4}=t$
We have,
$$\,dt = \frac{4(x^{-3}-x^{-5})}{2\sqrt{2-2x^{-2}+x^{-4}}}\,dx$$
$$I = \frac{1}{2}\int dt$$
$$I = \frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+C = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
A: $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}=\int \dfrac{1-\dfrac{1}{x^2}}{x^3 \sqrt{2-\dfrac{2}{x^2}+\dfrac{1}{x^4}}}dx$$$$=\int \frac{\left(1-\dfrac{1}{x^2}\right)dx}{x^3 \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\int \frac{2\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{-2}{x^3}\right)dx}{ \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\int \frac{d\left(\left(\dfrac{1}{x^2}-1\right)^2+1\right)}{ \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\cdot 2 \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}\ +C$$
$$=\color{blue}{\frac{1}{2x^2}\sqrt{2x^4-2x^2+1}}\ +C$$
A: Let $I=\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$. Divide Numerator and denominator of integrand by $x^2$ to get: $I=\int \frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}} \mathop{dx}$. Substitute $y=2-2x^{-2}+x^{-4}$ so that $dy=4(x^{-3}-x^{-5})dx$ and hence $I=\int \frac{1}{4 \sqrt{y}} \mathop{dy}=\frac{y^{1/2}}{2}+C=\frac{\sqrt{2-2x^{-2}+x^{-4}}}{2}+C$, where $C$ is integration constant.
