Integrating Arc Functions: \intx*arctg(1/x)dx

[Dell]

\intx*arctg(1/x)dx

i have such problems integrating arc functions, don't know why, but they never turn out right,,

by using
\intudv=uv-\intvdu

u=x
du=dx

dv=arctg(1/x)
v=[1/(1+(1/x)²)]*ln|x|
=[x²/1+x²]ln|x|

\intx*arctg(1/x)dx=x*arctg(1/x)-\int(x³ln|x|)/(x²+1)dx

now...

 

[Mark44]

Your work going from dv to v is incorrect.
You have dv = arctan(1/x) (minor quibble: it should be arctan(1/x)dx), but from that you can't get to v = [1/(1 + (1/x)^2)]*ln|x|.

A possibility when you're doing integration by parts is arriving at an integral that's harder than the one you started with. That's usually a sign that what you picked for u and dv is not the right choice.

For this problem, I believe that the right choice is

u = arctan(1/x), dv = xdx

 

[Dell]

arctan(1/x)=u
du=[1/(1 + (1/x)^2)]*ln|x|*dx

xdx=dv
v=0.5x^2

is this correct?
this will also give me an integration harder than i started with

\intx*arctg(1/x)dx=0.5x^2arctan(1/x)-\int[0.5x^2/(1 + (1/x)^2)]*ln|x|dx

 

[Mark44]

arctan(1/x)=u
du=[1/(1 + (1/x)^2)]*ln|x|*dx

xdx=dv
v=0.5x^2

is this correct?

For du, you're close, but you have the antiderivative of 1/x rather than its derivative. IOW, you have d/dx(1/x) = ln|x|, which is wrong.
Starting with u = arctan(1/x), and letting w = 1/x, we have
u = arctan(w).
So du = d/dw(arctan(w)) *dw = d/dw(arctan(w)) * dw/dx * dx
= 1/[1 + w^2] * (-1)/x^2 * dx (the factor before dx is where you made your mistake.)
= 1/[1 + (1/x)^2]* (-1)/x^2 * dx
\frac{-dx}{(1 + (1/x)^2)x^2}
= \frac{-dx}{x^2 + 1}

this will also give me an integration harder than i started with

\intx*arctg(1/x)dx=0.5x^2arctan(1/x)-\int[0.5x^2/(1 + (1/x)^2)]*ln|x|dx