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Question Number 7030 by Tawakalitu. last updated on 07/Aug/16

$$\int \frac{2x^2}{\sqrt{1-x^4}}dx$$

Answered by uchechukwu okorie favour last updated on 10/Aug/16

$$letu=\sqrt{1-x^4}$$$$\frac{du}{dx}=-\frac{2x^3}{\sqrt{1-x^4}}$$$$\frac{dx}{du}=-\frac{\sqrt{1-x^4}}{2x}$$$$\Rightarrow \int \frac{2x^3}{\sqrt{1-x^4}}dx=\int \frac{2x^3}{\sqrt{1-x^4}}.\frac{dx}{du}.\frac{du}{dx}.dx$$$$\Rightarrow \int \frac{2x^3}{\sqrt{1-x^4}}dx=\int \frac{2x^3}{\sqrt{1-x^4}}.-\frac{\sqrt{1-x^4}}{2x}$$$$=-\frac{\sqrt{1-x^4}}{x}\int \frac{1}{\sqrt{1-x^4}}$$$$=-\frac{\sqrt{1-x^4}}{x}In\mid 1-x^4\mid$$$$$$$$$$$$$$$$$$$$$$

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