How do you integrate int xtan^-1x by integration by parts method?
1 Answer
I got:
(x^2)/2arctanx + arctanx/2 - x/2 + C
Just pick the term that is easier to integrate, and use that as your
For integration by parts, you always start with:
\mathbf(int udv = uv - intvdu)
So, we can pick our
Then, of course, now that we've done that,
u = arctanx, du = 1/(1+x^2)dx
dv = xdx, v = x^2/2
Therefore, just follow the formula:
color(blue)(int xarctanxdx)
= (x^2)/2arctanx - 1/2intx^2/(1+x^2)dx
Now the challenge is to integrate this one. Fortunately, it's not that bad.
=> (x^2)/2arctanx - 1/2[int (x^2 + 1 - 1)/(x^2 + 1)dx]
= (x^2)/2arctanx - 1/2[int dx - int 1/(x^2 + 1)dx]
= (x^2)/2arctanx - 1/2[x - arctanx]
And finishing things up:
= color(blue)((x^2)/2arctanx + arctanx/2 - x/2 + C)
