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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.

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5 Answers 5

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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.

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

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With Euler Sub$\ldots$ $$ x = {\root{\root{2} - \root{2}t^{2}} \over \root{2}\root{\root{2} + 2t}} \quad\imp\quad t \equiv \root{2x^{4} - 2x^{2} + 1} - \root{2}x^{2} $$ the integral adopt a relatively simple form \begin{align} \int{x^{2} - 1 \over x^{3}\root{2x^{4} - 2x^{2} + 1}}\,\dd x & = \int{t^{2} + 2\root{2}t + 1 \over \pars{t^{2} - 1}^{2}}\,\dd t \\[3mm] & = \int{\dd t \over t^{2} - 1}\,\dd t + \root{2}\int{2t\,\dd t \over t^{2} - 1}\,\dd t + 2\int{\dd t \over \pars{t^{2} - 1}^{2}} = -\,{t + \root{2} \over t^{2} - 1} \end{align}

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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 $$I=\frac{1}{\sqrt{2}(\sin v + \cos v)}$$ $$=\frac{1}{\sqrt{2}(\sin(\arctan u)+\cos(\arctan u))}=\frac{\sqrt{u^2+1}}{\sqrt{2}\left(u + 1\right)}$$ $$=\frac{\sqrt{4 x^4-4 x^2+2}}{2 \sqrt{2}x^2}=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}$$

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

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