MHB Can you help me solve this tricky integral involving arctanx?

[DanielBW]

Hello everyone! I need some help with the integral:

$\displaystyle \int \dfrac{1}{\tan^{-1}(x)}dx$

I don't know how to solve it... can you guys help me please?

 

[MarkFL]

That function (the integrand) does not have an anti-derivative that can be expressed in elementary terms (at least according to W|A). Is this perhaps part of another problem, where this may be the wrong result?

 

[DanielBW]

That function (the integrand) does not have an anti-derivative that can be expressed in elementary terms (at least according to W|A). Is this perhaps part of another problem, where this may be the wrong result?

Well actually... yes, the orginal problem is to demostrate that the differential equation $xdy-ydx=tan^{-1}(y/x)dx$ can be solved by using the substitution $y=vx$ even for this non-homogeneus equation. So i proceed to solve:

$y=vx$

Then

$dy=vdx+xdv$

Substituing in the original differential equation i did obtain:

$x(vdx+xdv)-vxdx=tan^{-1}(vx/x)dx$

Simplifiyng and re-ordening the equation i get:

$x^{2}dv-{tan}^{-1}(v)dx=0$ (Separable differential equation)

$\dfrac{1}{tan^{-1}(v)}dv=\dfrac{1}{x^2}dx$

Then... to solve it:

$\int \dfrac{1}{tan^{-1}(v)}dv=\int \dfrac{1}{x^2}dx$

$\int \dfrac{1}{tan^{-1}(v)}dv=- \dfrac{1}{x}+C$

But to complete the problem i need to solve the integral and return to the original variable $y$.

 

[Siron]

I can't see any mistake in your solution so I guess you can just leave it like that and argue that de differential equation has only an implicit solution.

 

[topsquark]

Or maybe that y = vx doesn't work after all.

-Dan