# Need help on integration question i found on net

1. Sep 26, 2009

### Keval

Question is in orange, answer is in black.

I got no idea how they got this answer :\

The way im trying is using a substition of $kx^2 = gsin\theta$

Just the one question i cant get my head aroung in this exercise step by step method would be appreciated

2. Sep 26, 2009

### tiny-tim

Hi Keval!

Not sin, but tanh … try substituting x√(k/g) = tanhu

3. Sep 26, 2009

### g_edgar

With denominator instead k x^2 + g you get an arctan answer, right? Use the same steps on this one and get an atanh answer.

4. Sep 26, 2009

### Keval

someone check this out for me ??

$-\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}$

to simplify i let$\frac{g}{k}=m^2$ to get

$-\frac{1}{k}\int \frac{dx}{x^2-m^2}=-\frac{1}{k}\int \frac{dx}{(x-m)(x+m)}$

Using partial fractions i got
$\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=$

$\frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=$

$-\frac{1}{m}\cdot \frac{1}{2}\ln \left|\frac{x+m}{x-m} \right|=-\frac{1}{m}\tanh^{-1}\left( \frac{x}{m}\right)$

$\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)$

5. Sep 27, 2009

### tiny-tim

Hi Keval!

Yes, your partial fractions method is fine.

But it would be far quicker to start "let x√(k/g) = tanhu, dx√(k/g) = sech2u du" …

try it!