General Calculus.

**dadon** · April 24th 2006, 08:49 AM

Hi all!

I need help on two calculus problems:

Firstly ...

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

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

What are the rules to get this problem solved?

Secondly ...

When doing Intergration on partial fraction lets take an example:

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

Then the integral becomes

$<br /> \int3ln(x+2) + ln(x-3) ... ?$

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

I hope this will help others too!

**ThePerfectHacker** · April 24th 2006, 01:53 PM

Originally Posted by dadon

Hi all!

I need help on two calculus problems:

Firstly ...

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

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

This is chain rule.
You got,
$y=\tan^{-1}(u)$ where $u=\ln x$
Thus,
$\frac{dy}{du}=\frac{1}{1+u^2}$ and
$\frac{du}{dx}=\frac{1}{x}$
Thus,
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
Thus,
$\frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{1+u^2}$
But, $u=\ln x$ thus,
$\frac{1}{x}\cdot \frac{1}{1+\ln^2 x}$

**ThePerfectHacker** · April 24th 2006, 01:59 PM

Originally Posted by dadon

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

You need,
$\int \frac{5}{(x-3)^2}dx$
Let, $u=x-3$
Then, $du=dx$
Thus, use this substitution to get,
$\int 5u^{-2}du=-5u^{-1}+C$
But, $u=x-3$ thus,
$-5(x-3)^{-1}+C$

**dadon** · April 25th 2006, 06:06 AM

Thank you! I appreciate your help and nicely explained!

Regards,

dadon

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