# Math Help - integration

1. ## integration

How do you integrate 1/(2v + 1)^2?

2. 2v+1 = u
du=2dv

and so on ...

u must get :

$\displaystyle \int \frac {1}{(2v+1)^2} \;dv= -\frac {1}{4v+2}+ C$

How do you integrate 1/(2v + 1)^2?
Let $u = 2v+1$, then $\dfrac{du}{dv} = 2 \Rightarrow dv = \dfrac{du}{2}$. Thus $\displaystyle \int\frac{1}{(2v+1)^2}\;{dv} = \frac{1}{2}\int\frac{1}{u^2}\;{du}$.
Continue from there.

4. Use a $u$ substitution.

$\int{\frac{1}{(2v + 1)^2}\,dv} = \frac{1}{2}\int{2(2v+1)^{-2}\,dv}$.

Now let $u=2v + 1$ so that $\frac{du}{dv} = 2$ and the integral becomes

$\frac{1}{2}\int{u^{-2}\,\frac{du}{dv}\,dv}$

$= \frac{1}{2}\int{u^{-2}\,du}$

$= \frac{1}{2}\left(\frac{u^{-1}}{-1}\right) + C$

$= -\frac{1}{2u} + C$

$= -\frac{1}{2(2v+1)} + C$

$= -\frac{1}{4v+2} + C$.