Thread: [SOLVED] Stuck on improper fractions in integrals

1. [SOLVED] Stuck on improper fractions in integrals

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...

2. 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.

3. $\displaystyle \frac{y^2}{1+25y^2} = \frac{1}{25}\left(1 - \frac{1}{1+25y^2}\right)$

4. Equivalently: $\displaystyle \int \frac{y^2}{1+25y^2} \ dy \ = \ \frac{1}{25} \int \frac{25y^2 {\color{red}+ 1 - 1}}{1 + 25y^2} \ dy \ = \ \frac{1}{25}\int \left(\frac{25y^2 + 1}{1+25y^2} - \frac{1}{1+(5y)^2}\right) \ dy$

5. $\displaystyle \int \frac{y^2}{1+25y^2}dy$

$\displaystyle tan(u)=5y$

$\displaystyle y=\frac{tan(u)}{5}$

$\displaystyle dy=\frac{sec^2(u)}{5}du$

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 $\displaystyle \frac{1}{125}\bigg(5y - arctan(5y)\bigg)$

6. 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:

$\displaystyle \frac {1}{25}\int 1 - \frac {1}{25}\int \frac {1}{1+25y^2}$

7. 5y-arctan(5y)

Is this a solution to the original integral?
$\displaystyle \int \frac {y^2}{1+25y^2}~dy$

I don't see exactly how that is the solution. I am stuck on the step before when you split the integral.

8. Originally Posted by redman223
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:

$\displaystyle \frac {1}{25}\int 1 - \frac {1}{25}\int \frac {1}{1+25y^2}$
$\displaystyle \int \frac {1}{25} (1-\frac {1}{1+25y^2}) ~dy$
$\displaystyle =\frac 1{25} \left(\int 1 ~dy - \int \frac{1}{1+25y^2} ~dy\right)=\frac 1{25} \int 1 ~dy-\frac 1{25} \int \frac{1}{1+25y^2} ~dy$