Find:\(\int\cfrac{x^2}{1-x^2}dx\)

1.	Find:\(\int\cfrac{x^2}{1-x^2}dx\)
Answer» \(\int\cfrac{x^2}{1-x^2}dx\) = \(\int\cfrac{1-1+x^2}{1-x^2}dx\) = \(\int\cfrac{1-(1+x^2)}{1-x^2}dx\) = \(\int\cfrac{1}{1-x^2}dx\) - \(\int dx\) = \(\cfrac12\)ln \(\left\|\cfrac{h\,x}{1-x}\right\|\)- x + c

Answer»

\(\int\cfrac{x^2}{1-x^2}dx\)

= \(\int\cfrac{1-1+x^2}{1-x^2}dx\)

= \(\int\cfrac{1-(1+x^2)}{1-x^2}dx\)

= \(\int\cfrac{1}{1-x^2}dx\) - \(\int dx\)

= \(\cfrac12\)ln \(\left|\cfrac{h\,x}{1-x}\right|\)- x + c

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Find:\(\int\cfrac{x^2}{1-x^2}dx\)

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