Question

Question #c1742

Answer 1

(2)

Explanation:

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`