How would one go about integrating this? Please provide full working.
$\displaystyle \int\frac{y}{(y^2+3)} dy$
My math teacher told me to let $\displaystyle u = y^2+3$ and use the derivative of that, but I still don't quite understand what to do.
How would one go about integrating this? Please provide full working.
$\displaystyle \int\frac{y}{(y^2+3)} dy$
My math teacher told me to let $\displaystyle u = y^2+3$ and use the derivative of that, but I still don't quite understand what to do.
Rewrite it as
$\displaystyle \int{(y^2 + 3)^{-1}y\,dy}$
$\displaystyle = \frac{1}{2}\int{(y^2 + 3)^{-1}\,2y\,dy}$.
Now let $\displaystyle u = y^2 + 3$ so that $\displaystyle \frac{du}{dy} = 2y$.
The integral becomes
$\displaystyle \frac{1}{2}\int{u^{-1}\,\frac{du}{dy}\,dy}$
$\displaystyle = \frac{1}{2}\int{u^{-1}\,du}$
$\displaystyle = \frac{1}{2}\ln{|u|} + C$
$\displaystyle = \frac{1}{2}\ln{|y^2 + 3|} + C$
$\displaystyle = \frac{1}{2}\ln{(y^2 + 3)} + C$ since $\displaystyle y^2 + 3 > 0$ for all $\displaystyle y$.
You can just let $\displaystyle u = y^2+3$ straight away (without creating the derivative of the denominator in the numerator): if $\displaystyle u = y^2+3$, then $\displaystyle \dfrac{du}{dy} = 2u \Rightarrow dy = \dfrac{du}{2y}$. That means: $\displaystyle \int\dfrac{y}{y^3+3}\;{dy}= \int\dfrac{y}{\left(u\right)2y}\;{du} = \dfrac{1}{2} \int\dfrac{y}{\left(u\right)y}\;{du} = \dfrac{1}{2} \int\dfrac{1}{\left(u\right)}\;{du} = \dfrac{1}{2}\log{u}+C$ $\displaystyle \Rightarrow \int\dfrac{y}{y^3+3}\;{dy} = \dfrac{1}{2}\log\left(y^3+3\right)+C$