1. ## How to integrate..

$\int\frac{y}{(y^2+3)} dy$

My math teacher told me to let $u = y^2+3$ and use the derivative of that, but I still don't quite understand what to do.

2. Originally Posted by Exotique

$\int\frac{y}{(y^2+3)} dy$

My math teacher told me to let $u = y^2+3$ and use the derivative of that, but I still don't quite understand what to do.
Rewrite it as

$\int{(y^2 + 3)^{-1}y\,dy}$

$= \frac{1}{2}\int{(y^2 + 3)^{-1}\,2y\,dy}$.

Now let $u = y^2 + 3$ so that $\frac{du}{dy} = 2y$.

The integral becomes

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

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

$= \frac{1}{2}\ln{|u|} + C$

$= \frac{1}{2}\ln{|y^2 + 3|} + C$

$= \frac{1}{2}\ln{(y^2 + 3)} + C$ since $y^2 + 3 > 0$ for all $y$.

3. Originally Posted by Exotique
$\int\frac{y}{(y^2+3)} dy$
My math teacher told me to let $u = y^2+3$ and use the derivative of that, but I still don't quite understand what to do.
You can just let $u = y^2+3$ straight away (without creating the derivative of the denominator in the numerator): if $u = y^2+3$, then $\dfrac{du}{dy} = 2u \Rightarrow dy = \dfrac{du}{2y}$. That means: $\int\dfrac{y}{y^3+3}\;{dy}= \int\dfrac{y}{\left(u\right)2y}\;{du} = \dfrac{1}{2} \int\dfrac{y}{\left(u\right)y}\;{du} = \dfrac{1}{2} \int\dfrac{1}{\left(u\right)}\;{du} = \dfrac{1}{2}\log{u}+C$ $\Rightarrow \int\dfrac{y}{y^3+3}\;{dy} = \dfrac{1}{2}\log\left(y^3+3\right)+C$