Indefinite integral involving arctan and ln

[appplejack]

Homework Statement

∫ 1/x arctan (lnx) dx

Homework Equations

The Attempt at a Solution

1.U substitution. SO u = ln x, du= 1/x dx
∫ arctan u du

2.by parts: u = arctan u du = 1/ 1+u^2
v = 1 dv = du

3. uv - ∫vdu = artan u - ∫ 1/ 1+u^2
= arctan u - artan u = 0

Did I answer this right?

 

[Dick]

If du=dv, then v=u, not v=1! I'd also suggest when you get to ∫ arctan u du and if you want to use u and v for the variables in the integration by parts you change that to ∫ arctan w dw. Otherwise the variable naming gets really confusing.

 

[scurty]

Also, u = arctan(u) du doesn't make sense! If you already did u-substitution, use v and w for integration by parts so your variables don't get mixed up.

 

[Dick]

Also, u = arctan(u) du doesn't make sense! If you already did u-substitution, use v and w for integration by parts so your variables don't get mixed up.

I think applejack meant u=arctan(u), which also doesn't make sense, and is part of the variable naming problem.

 

[scurty]

I think applejack meant u=arctan(u), which also doesn't make sense, and is part of the variable naming problem.

You're right, he did, the du part is when he took the derivative.

I'm used to putting the four equations (u, dv, du, v) in a box shape in my scratch work so I always get confused when it is formatted poorly on websites. Regardless, in addition to changing the variable, there should have been a comma inserted there.