How do you find the integral of #xarctanxdx#?

Question

Charles Anctil · Accepted Answer

As you can see, there is one element that is simple to integrate and one that is feasible to separate. Although you might already know #int arctanxdx#, I will assume that you don't. Instead, I will assume that you know #d/(dx)[arctanx] = 1/(1+x^2)#. This is what I see when I look at it: #int xarctanxdx = int udv# which implies Parts Integration. #int udv = uv - int vdu# Let:
#u = arctanx#
#du = 1/(1+x^2)dx#
#dv = xdx#
#v = x^2/2# #=> x^2/2 arctanx - 1/2int x^2/(1+x^2)dx# (One might be tempted to use partial fraction decomposition with the integral here, but there is a simpler method.) #1/2 int x^2/(1+x^2)dx = 1/2 int (1 + x^2 - 1)/(1+x^2)dx# #= 1/2 int 1 - 1/(1+x^2)dx# #= 1/2(x - arctanx)# Consequently, the total integral yields: #int xarctanxdx = x^2/2 arctanx - 1/2(x - arctanx)# #= color(blue)(1/2[(x^2 + 1)arctanx - x] + C)#

Elijah Abram · Answer

undefined To find the integral of $x\arctan(x)$, you can use integration by parts. Let $u = \arctan(x)$ and $dv = x \, dx$. Then, differentiate $u$ to find $du$ and integrate $dv$ to find $v$. After that, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values of $u$, $dv$, $du$, and $v$, and then integrate. You will eventually get the result.