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!