18

Dear student 

This integral can be evaluated as follows
 $$\begin{gathered} Integral, I=\int \frac{x^2}{1-x^4}dx=\int \frac{x^2}{1-{\left(x^2\right)}^2}dx \\ \Rightarrow I=\int \frac{x^2}{1-{\left(x^2\right)}^2}dx=\int \frac{x^2}{(1+x^2)(1-x^2)}dx \\ \Rightarrow I=\int \frac{x^2}{(1+x^2)(1-x^2)}dx=\int \frac{1-(1-x^2)}{(1+x^2)(1-x^2)}dx \\ \Rightarrow I=\int \frac{1-(1-x^2)}{(1+x^2)(1-x^2)}dx=\int \frac{1}{(1+x^2)(1-x^2)}dx-\int \frac{(1-x^2)}{(1+x^2)(1-x^2)}dx \\ \Rightarrow I=\int \frac{1}{(1+x^2)(1-x^2)}dx-\int \frac{1}{(1+x^2)}dx=\int \frac{1}{(1-x^4)}dx-\int \frac{1}{(1+x^2)}dx \\ \Rightarrow I=\int \frac{1}{(1-x^4)}dx-\int \frac{1}{(1+x^2)}dx \\ let I=I_1-I_2 \\ {I}_2=\int \frac{1}{(1+x^2)}dx={\tan }^{-1}\left(x\right)+ constant\mathit{ } \\ to\mathit{ }evaluate\mathit{ }{I}_{\mathit{2}}\mathit{ }the\mathit{ }above\mathit{ }procedure\mathit{ }is\mathit{ }again\mathit{ }repeated\mathit{ } \\ we\mathit{ }use\mathit{ }the\mathit{ }equation\mathit{ } \\ \int \frac{1}{(1-x^2)}dx=\frac{1}{2}\log \left|\frac{1+x}{1-x}\right|+ C, then we get \\ {I}_1=\int \frac{1}{(1-x^4)}dx=\frac{1}{4}\left(-\log (1-x\right)+\log (1+x)+2{\tan }^{-1}\left(x\right))+C \\ Thus \\ I=I_1-I_2 \\ \Rightarrow I=\int \frac{1}{(1-x^4)}dx-\int \frac{1}{(1+x^2)}dx \\ \Rightarrow I\mathit{=}\frac{1}{4}\left(-\log (1-\mathrm{x}\right)+\log (1+\mathrm{x})+2{\tan }^{-1}(\mathrm{x}\left)\right)-{\tan }^{-1}\left(\mathrm{x}\right) \\ \Rightarrow \mathrm{I}=\frac{1}{4}\left(-\log (1-\mathrm{x}\right)+\log (1+\mathrm{x})-2{\tan }^{-1}\left(\mathrm{x}\right))+\mathrm{C} \\ \int \frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^4}\mathrm{dx}=\frac{1}{4}\left(-\log (1-\mathrm{x}\right)+\log (1+\mathrm{x})-2{\tan }^{-1}(\mathrm{x}\left)\right)+\mathrm{C} \end{gathered}$$

Regards