# How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ [closed]

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

I tried using the substitution $x^2=z$.But that did not help much.

You can apply $u=x^2$ : $$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx=\frac{1}{2}\int \frac{x^2-1}{x^4\sqrt{2x^4-2x^2+1}} 2xdx=\frac{1}{2}\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du$$ Now note that : $$\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du=\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du$$ we can substitute $v=\frac{1}{u}$ : $$\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du=\int \frac{1-u}{\sqrt{2(u^2-u)+1}} \frac{-du}{u^2}=\int \frac{1-\frac{1}{v}}{\sqrt{2(\frac{1}{v^2}-\frac{1}{v})+1}} dv\\=\int \frac{\frac{v-1}{v}}{\sqrt{\frac{2-2v+v^2}{v^2}}}dv=\int \frac{v-1}{\sqrt{v^2-2v+2}}dv$$ Substitute $y=v^2-2v+2$ and you should be able to finish.

By setting $x^2=z$ we are left with: $$\frac{1}{2}\int\frac{z-1}{z^2\sqrt{2z^2-2z+1}}\,dz=C+\frac{\sqrt{2z^2-2z+1}}{2z}.$$


A very long way... (Let $I$ equal the integral) $$I=\frac{1}{\sqrt{2}}\int\frac{x^2-1}{x^3\sqrt{(x^2-1/2)^2+1/4}}dx$$ Let $(x^2-1/2)=u/2$ meaning $x=\sqrt{1/2(u+1)}$ and $dx=\frac{1}{4\sqrt{1/2(u+1)}}$ Hence $$I=\frac{1}{\sqrt{2}}\int\frac{u-1}{(u+1)^2\sqrt{u^2+1}}du$$ Then let $u=\tan(v)$ and $du=\sec^2(v)$ giving $$I=\frac{1}{\sqrt{2}}\int\frac{(\tan v -1)\sec v}{(\tan v+1)^2}dv$$ Using $\tan v=\frac{\sin v}{\cos v}$ and $\sec v=\frac{1}{\cos v}$ we have $$I=\frac{1}{\sqrt{2}}\int\frac{\sin v-\cos v}{(\sin v +\cos v)^2}dv$$ Finally this can be evaluated using $m=\sin v + \cos v$ and $dm=\cos v - \sin v$ to get $$I=-\frac{1}{\sqrt{2}}\int\frac{1}{m^2}dm=\frac{1}{\sqrt{2}m}$$ Doing substitutions back up to $x$ we have $$I=\frac{1}{\sqrt{2}(\sin v + \cos v)}$$ $$=\frac{1}{\sqrt{2}(\sin(\arctan u)+\cos(\arctan u))}=\frac{\sqrt{u^2+1}}{\sqrt{2}\left(u + 1\right)}$$ $$=\frac{\sqrt{4 x^4-4 x^2+2}}{2 \sqrt{2}x^2}=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}$$

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

So $$I=\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$

Now Put $2-2x^{-2}+x^{-4} = t^2\;,$ Then $4(x^{-3}-x^{-5})dx = 2tdt$

So $$I = \frac{1}{2}\int\frac{t}{t}dt = \frac{1}{2}t+\mathcal{C}=\frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+\mathcal{C}$$

So $$I = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+\mathcal{C}$$

• Tell me something.How did you think of that step where you took $x^2$ common out? It is literally out of the box!
– user220382
Jul 3, 2016 at 6:16
• In these type of method, We generally take highest power of $x$ common from Denominator. Jul 4, 2016 at 7:37