Q 20 ∫ x 2 1 - x 4 d x

Hi, ∫x21-x4dx=12∫2x21-x4dx2x21-x4=2x21-x21+x2=11-x2-11+x2=11-x1+x-11+x2=1211+x+11-x-11+x2putting we get =12∫1211+x+11-xdx-12∫11+x2dx=14ln1+x-ln1-x-12tan-1x+C= 14 ln 1+x1-x-12tan-1x+C

Q.20. $\int \frac{x^{2}}{1 - x^{4}} d x$

Hi, \int \frac{x^{2}}{1 - x^{4}} dx = \frac{1}{2} \int \frac{2 x^{2}}{1 - x^{4}} dx \frac{2 x^{2}}{1 - x^{4}} = \frac{2 x^{2}}{(1 - x^{2}) (1 + x^{2})} = \frac{1}{1 - x^{2}} - \frac{1}{1 + x^{2}} = \frac{1}{(1 - x) (1 + x)} - \frac{1}{1 + x^{2}} = \frac{1}{2} [\frac{1}{(1 + x)} + \frac{1}{(1 - x)}] - \frac{1}{1 + x^{2}} putting we get = \frac{1}{2} \int \frac{1}{2} [\frac{1}{(1 + x)} + \frac{1}{(1 - x)}] dx - \frac{1}{2} \int \frac{1}{1 + x^{2}} dx = \frac{1}{4} [\ln (1 + x) - \ln (1 - x)] - \frac{1}{2} \tan^{- 1} x + C = \frac{1}{4} \ln \frac{1 + x}{1 - x} - \frac{1}{2} \tan^{- 1} x + C

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Q.20. ∫ x 2 1 - x 4 d x

Q.20. $\int \frac{x^{2}}{1 - x^{4}} d x$