What is #int (ln(3x-1))^-2dx#?

1 Answer
May 21, 2018

There's no known results of this equation.

Explanation:

#int1/(ln(3x-1)²)dx#

Let #X=3x-1#

#dX=3dx#
So:
#int1/(ln(3x-1)²)dx=1/3int1/(ln(X)²)dX#
Now let #Y=ln(X)#
#X=e^Y#
#dX=e^YdY#
So:

#1/3int1/(ln(X)²)dX=1/3inte^Y/(Y²)dY#
Using integration by parts :
#intf*g'dY=f*g-intg*f'dY#
There:
#f(Y)=e^Y#, #g'(Y)=1/(Y²)#
#f'(Y)=e^Y#, #g(Y)=-1/Y#

#1/3inte^Y/(Y²)dY=1/3(-e^Y/Y)+1/3inte^Y/YdY#

But because of Liouville's theorem, there's no known solutions for #inte^Y/YdY#
\0/ here's our answer !