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Question Number 126604 by bramlexs22 last updated on 22/Dec/20

$$\int \frac{dx}{(1-x^2)\sqrt[4]{2x^2-1}}?$$

Answered by liberty last updated on 22/Dec/20

$$Y=\int \frac{(1-x^2)}{{(1-x^2)}^2\sqrt[4]{2x^2-1}}dx$$$$Y=\int \frac{(1-x^2)}{(x^4-2x^2+1)\sqrt[4]{2x^2-1}}dx$$$$Y=\int \frac{(1-x^2)}{(x^4-(2x^2-1))\sqrt[4]{2x^2-1}}dx$$$$let\frac{x}{\sqrt[4]{2x^2-1}}=w;dw=\frac{x^2-1}{(2x^2-1)\sqrt[4]{2x^2-1}}dx$$$$Y=\int \frac{dw}{1-w^4}=\frac{1}{2}\int \frac{dw}{1+w^2}+\frac{1}{2}\int \frac{dw}{1-w^2}$$$$Y=\frac{1}{2}{\tan }^{-1}\left(w\right)+\frac{1}{2}(\frac{1}{2}\ln \mid \frac{1+w}{1-w}\mid )+c$$$$Y=\frac{1}{2}{\tan }^{-1}\left(\frac{x}{\sqrt[4]{2x^2-1}}\right)+\frac{1}{4}\ln \mid \frac{\sqrt[4]{2x^2-1}+x}{\sqrt[4]{2x^2-1}-x}\mid +c$$$$$$

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