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Evaluate: `int(x^2tan^(-1)x)/(1+x^2) dx` |
Answer» Let `x=tan theta.` ` :. dx=sec^(2) theta d theta` ` :. I=int (x^(2)tan^(-1)x)/(1+x^(2))dx` `=int (theta tan^(2) theta)/(1+tan^(2)theta)*sec^(2) theta d theta` `= int theta tan^(2) theta d theta` `=int theta (sec^(2) theta-1) d theta` `=int theta sec^(2) theta d theta - int theta d theta` `=theta tan theta -int 1*tan theta d theta-(theta^(2))/(2) +c` `=theta tan theta -log_(e)|cos theta|-(theta^(2))/(2)+c, " where " theta = tan^(-1)x`. |
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