Question #c1742

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Answer 1 · 2016-05-22T21:26:09

(2)

Distribute $e^(tan^-1(x))$ in the integrand and split up the integrand through addition:

$=inte^(tan^-1(x))dx+int(xe^(tan^-1(x)))/(1+x^2)dx$

Attempt to solve the $inte^(tan^-1(x))dx$ through integration by parts:

$intudv=uv-intvdu$

Let $u=e^(tan^-1(x))$ and $dv=dx$ . These imply that $du=e^(tan^-1(x))/(1+x^2)dx$ and $v=x$ .

Thus, $inte^(tan^-1(x))dx=xe^(tan^-1(x))-int(xe^(tan^-1(x)))/(1+x^2)dx$

When we plug this back into the original expression, we see that the $int(xe^(tan^-1(x)))/(1+x^2)dx$ terms will cancel one another.

$=xe^(tan^-1(x))-int(xe^(tan^-1(x)))/(1+x^2)dx+int(xe^(tan^-1(x)))/(1+x^2)dx$

$=xe^(tan^-1(x))+C$