# Thread: Showing this equals that

1. ## Showing this equals that

Sorry for the vague title, it's not on a specific topic, it's a section in a question and differential equations and I've been told to show that...

$\frac{1}{n(1-\lambda n^{\alpha})}=\frac{1}{n}+\frac{\lambda n^{\alpha -1}}{1-\lambda n^{\alpha}}$

...then I've got to integrate the RHS. I know how to integrate that, but I'm stuck on how to show LHS = RHS. Any ideas?

2. Hello,
Originally Posted by chella182
Sorry for the vague title, it's not on a specific topic, it's a section in a question and differential equations and I've been told to show that...

$\frac{1}{n(1-\lambda n^{\alpha})}=\frac{1}{n}+\frac{\lambda n^{\alpha -1}}{1-\lambda n^{\alpha}}$

...then I've got to integrate the RHS. I know how to integrate that, but I'm stuck on how to show LHS = RHS. Any ideas?
It's like a partial fractions decomposition :
Solve for a and b in $\frac{1}{n(1-\lambda n^\alpha)}=\frac an+\frac{b}{1-\lambda n^\alpha}$
(by putting on the same fraction)

Otherwise, you can start from the RHS to prove the LHS.

Another method is to note that $1=(1-\lambda n^\alpha)+\lambda n^\alpha$

3. I tried partial fractions and got in a right mess with it :\.

4. Okay, proved the thing... I can't figure how to integrate

$\frac{\lambda n^{\alpha-1}}{1-\lambda n^{\alpha}}$

though :\.