Integration by parts!! I am stuck!!

Hi everyone!

So I was working on this question and we are using integration by parts.

integral of arctan*(1/x) dx from sqrt(3) to 1.

I chose dv=1/x and v=lnx u=arctan and du=1/(x^2+1)dx.

The question is, how can the derivative of arctan be 1/(x^2+1)???? Can anyone please explain it to me because I am really confused!

Thanks!

First how can you have arctan*(1/x)
Do you mean arctan(1/x) OR arctan(x) * 1/x

Anyway to answer your question:
y = arctan(x)
tany = x

Differentiate both sides with respect to x
sec²y dy/dx = 1
(tan²y + 1) dy/dx = 1 . . . . . . . where tany = x
(x² + 1) dy/dx = 1
dy/dx = 1 / (x² + 1) . . . . . . . . where y = arctan(x)
d/dx (arctan(x)) = 1 / (x² + 1)

--------------------------------

To find ∫ arctan(x) * 1/x dx, using integration by parts as requested:
u = arctan(x) . . . . . . dv = 1/x dx
du = 1/(x²+1) dx . . . . v = lnx

∫ arctan(x) * 1/x dx = lnx arctan(x) − ∫ lnx/(x²+1) dx

But this doesn't get you any further.
In actuality, you can't find ∫ arctan(x) * 1/x dx in terms of regular functions:
http://www.wolframalpha.com/input/?i=%E2…

So again I'm left wondering if perhaps you really meant ∫ arctan(1/x) dx
instead of ∫ arctan(x) * 1/x dx

——————————————————————————————

I'll show you how to find ∫ arctan(1/x) dx using integration by parts:
∫ u dv = u v − ∫ v du

Note that in equation above, u and dv are multiplied, they are not composite functions. So we cannot separate arctan from 1/x. So we use the following substitutions:

u = arctan(1/x) ....... dv = dx

du/dx = 1/((1/x)²+1) * d/dx (1/x) = 1/(1/x² + 1) * -1/x² = −1/(1+x²)
du = −1/(x²+1) dx

v = x

∫ u dv = u v − ∫ v du
∫ arctan(1/x) dx = x arctan(1/x) − ∫ x * −1/(x²+1) dx
∫ arctan(1/x) dx = x arctan(1/x) + ∫ x/(x²+1) dx
∫ arctan(1/x) dx = x arctan(1/x) + 1/2 ln(x²+1) + C

Integrating from 1 to √3 we get
√3 arctan(1/√3) + 1/2 ln(3+1) − 1 arctan(1) − 1/2 ln(2)
= √3 π/6 + 1/2 ln(4) − π/4 − 1/2 ln(2)
= √3 π/6 − π/4 + 1/2 * 2 ln(2) − 1/2 ln(2)
= (2√3−3)π/12 + 1/2 ln(2)
= 0.468075109