Evaluate $$ \int{\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}}dx $$ I substituted $x- \frac{1}{x} $ with u so $$-\int{\frac{du}{u\sqrt{u^2+3}}} $$ Now I put $u=\frac1t$ so $$\int{\frac{dt}{\sqrt{1+3t^2}}} $$
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Related : http://math.stackexchange.com/questions/692118/integrate-fracx2-1x21-frac1-sqrt1x4dx – lab bhattacharjee Aug 12 '15 at 12:30
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HINT:
$$\text{Set }\int\left(1+\dfrac1{x^2}\right)dx=u\text{ so that }\dfrac{du}{dx}=\cdots$$
$$\text{in }\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)\sqrt{\dfrac1{x^2}+1+x^2}}$$ and use $$\dfrac1{x^2}+x^2=\left(x-\dfrac1x\right)^2+2$$
For $\int\dfrac{du}{u\sqrt{u^2+3}}$
either set $u=\sqrt3\tan y$
or for $\int\dfrac{du}{u\sqrt{u^2+3}}=\int\dfrac{u\ du}{u^2\sqrt{u^2+3}}$ set $\sqrt{u^2+3}=v$
lab bhattacharjee
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Yes I reached upto this step. I tried proceeding by making substitution $ u= \frac{1}{t} $. Can you help me on that ? – mathemather Aug 12 '15 at 06:34
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@arutoregni, Can you please update your answer with your method. Btw, I've suggested two other substitutions, right? – lab bhattacharjee Aug 12 '15 at 06:35