Question

Question #d8ccb

Answer 1

`(e^arctanx(x-1))/(2sqrt(1+x^2))+C`

Explanation:

`I=int(xe^arctanx)/((1+x^2)sqrt(1+x^2))dx`

Let `t=arctanx`. This implies that `dt=1/(1+x^2)dx` and that `x=tant`.

`I=int(tant(e^t))/sqrt(1+tan^2t)dt`

Recall that `1+tan^2t=sec^2t`:

`I=inte^t tant/sectdt=inte^tsintdt`

This is a tricky integral which is solved with two iterations of integration by parts. Start with:

`{(u=e^t,=>,du=e^tdt),(dv=sintdt,=>,v=-cost):}`

`I=-e^tcost+inte^tcostdt`

Now let:

`{(u=e^t,=>,du=e^tdt),(dv=costdt,=>,v=sint):}`

So:

`I=-e^tcost+e^tsint-inte^tsintdt`

The original integral `I=inte^tsintdt` has reappeared so we can add it to both sides of the equation then isolate `I`:

`2I=e^tsint-e^tcost`

`I=1/2e^t(sint-cost)`

Substituting back in with `t=arctanx`:

`I=1/2e^arctanx(sin(arctanx)-cos(arctanx))`

Note that the angle `arctanx` occurs in a right triangle where the side opposite the angle is `x` and the side adjacent to the angle is `1`. Through the Pythagorean Theorem, its hypotenuse is `sqrt(1+x^2)`.

Then, `sin(arctanx)=x/sqrt(1+x^2)` and `cos(arctanx)=1/sqrt(1+x^2)`.

`I=1/2e^arctanx(x/sqrt(1+x^2)-1/sqrt(1+x^2))`

`I=(e^arctanx(x-1))/(2sqrt(1+x^2))+C`