Evaluate : `int(tan^(-1)x)/((1+x)^(2))dx`.

1.	Evaluate : `int(tan^(-1)x)/((1+x)^(2))dx`.
Answer» Integrating by parts, taking `tan^(-1)`x as the first function and `(1)/((1+x)^(2))` as the second function, we get `I=tan^(-1)x((-1))/((1+x))-int(1)/((1+x^(2)))((-1))/((1+x))dx` `(-tan^(-1)x)/((1+x))+int(dx)/((1+x)(1+x^(2)))=(-tan^(-1)x)/((1+x))+(1)/(2)*int{(1)/((1-x))+((1-x))/((1+x^(2)))}" [by partial fractions]"` `=(-tan^(-1))/((1+x))+(1)/(2)log\|a+x\|+(1)/(2)tan^(-1)x-(1)/(4)log(1+x^(2))+C`.

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