Question

integral of x*arctan(x)dx

Answer 1

1/2 (-x+(1+x^2) tan^(-1)(x))+C

Answer 2

i need to know how to get there not just the answer I'm studying for a test not just trying to get homework done.

Answer 3

parts for this one but it is a pain

Answer 4

\[u=\tan^{-1}(x), du =\frac{1}{x^2+1}, dv=x, v=\frac{x^2}{2}\] gives
\[\frac{x^2}{2}\tan^{-1}(x)-\int \frac{x^2}{2(x^2+1)}dx\] but second integral is not so fun

Answer 5

right there is where I got stuck

Answer 6

@satellite73 please kick the spammer from the chat. Thanks

Answer 7

oh actually the second one is not so bad as i though, because
\[\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\] and the integral of the second piece is back to arctan

Answer 8

I don't see how \[x^{2}/x^{2}+1=1-1/x^{2}+1\]

Answer 9

divide

Answer 10

whenever the degree of the numerator is the same as the denominator (or larger) it is usually the right thing to do

Answer 11

i see it now thanks

Answer 12

or you can use a trick, namely
\[\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}\] but that trick really works best if you already know the answer

Answer 13

That is actually how I saw it in my head when it clicked