I am trying to integrate:
$\displaystyle
\int \frac {y^2}{1+25y^2}~dy
$
I am supposed to do long division right? I have come up with -1/25, but I don't know what to do with that number...
One thing you could do is hammer it into a form that is recognizable.
$\displaystyle \int\frac{y^{2}}{1+25y^{2}}dy$
Let $\displaystyle u=5y, \;\ \frac{du}{5}=dy$
When we make the subs we get:
$\displaystyle \frac{1}{125}\int\frac{u^{2}}{1+u^{2}}du$
$\displaystyle \frac{1}{125}\left[\int du-\int\frac{1}{u^{2}+1}du\right]$
See the right integral?. Do you notice something that has to do with arctan?.
Just integrate and resub the u=5y.
$\displaystyle \int \frac{y^2}{1+25y^2}dy$
$\displaystyle tan(u)=5y$
$\displaystyle y=\frac{tan(u)}{5}$
$\displaystyle dy=\frac{sec^2(u)}{5}du$
so $\displaystyle \int \frac{y^2}{1+25y^2}dy$ = $\displaystyle \int \frac {tan^2(u)}{125}du$ = $\displaystyle \frac{1}{125} \bigg(\int sec^2(u)du - \int 1du\bigg)$
which is $\displaystyle tan(u)-u$
I'll finish it up when I come back from tacobell, my girlfriend don't want to wait
Ok came to finish $\displaystyle \frac{1}{125}\bigg(5y - arctan(5y)\bigg)$
Ok thanks, this was just a small part of a larger problem, I just forgot what to do when it came to long division.
The way skeeter did it is what we learned in class the other day, but I am confused as far as what to do next.
$\displaystyle
\int \frac {1}{25} (1-\frac {1}{1+25y^2})
$
What do I do from here? Do I multiply the 1/25 by everything or do I pull the 1/25 out in front of the integral sign and create 2 more integrals like this:
$\displaystyle
\frac {1}{25}\int 1 - \frac {1}{25}\int \frac {1}{1+25y^2}
$
Please don't forget dy.
$\displaystyle \int \frac {1}{25} (1-\frac {1}{1+25y^2}) ~dy
$
Then you can pull out any constant you want, because the antiderivative of a*y is a*antiderivative of y.
$\displaystyle =\frac 1{25} \left(\int 1 ~dy - \int \frac{1}{1+25y^2} ~dy\right)=\frac 1{25} \int 1 ~dy-\frac 1{25} \int \frac{1}{1+25y^2} ~dy$
In the first one, calculate and then, divide by 25, properly.
In the second one, substitute carefully t=5y, because you did it wrong..