Hi all!

I need help on two calculus problems:

Firstly...

$\displaystyle y = tan^-^1(lnx)$

I know the solution which is $\displaystyle \frac{dy}{dx} = \frac{1}{x} . \frac{1}{1 + (lnx)^2}$

What are the rules to get this problem solved?

Secondly...

When doing Intergration on partial fraction lets take an example:

$\displaystyle

\int\frac{4x^2 - 24x + 11}{(x-3)^2(x+2)} = \frac{3}{(x+2)} + \frac{1}{(x-3)} - \frac{5}{(x-3)^2}$

Then the integral becomes

$\displaystyle

\int3ln(x+2) + ln(x-3) ... ?$

How can I integrate the last bit: $\displaystyle -\frac{5}{(x-3)^2} $

I hope this will help others too!