[SOLVED] Stuck on improper fractions in integrals

**redman223** · September 4th 2008, 04:30 PM

I am trying to integrate:

$ \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...

**galactus** · September 4th 2008, 05:08 PM

One thing you could do is hammer it into a form that is recognizable.

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

Let $u=5y, \;\ \frac{du}{5}=dy$

When we make the subs we get:

$\frac{1}{125}\int\frac{u^{2}}{1+u^{2}}du$

$\frac{1}{125}\left[\int du-\int\frac{1}{u^{2}+1}du\right]$

See the right integral?. Do you notice something that has to do with arctan?.

Just integrate and resub the u=5y.

**skeeter** · September 4th 2008, 05:11 PM

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

**o_O** · September 4th 2008, 05:12 PM

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

**11rdc11** · September 4th 2008, 05:44 PM

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

$tan(u)=5y$

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

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

so $\int \frac{y^2}{1+25y^2}dy$ = $\int \frac {tan^2(u)}{125}du$ = $\frac{1}{125} \bigg(\int sec^2(u)du - \int 1du\bigg)$

which is $tan(u)-u$

I'll finish it up when I come back from tacobell, my girlfriend don't want to wait

Ok came to finish $\frac{1}{125}\bigg(5y - arctan(5y)\bigg)$

**redman223** · September 4th 2008, 05:58 PM

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.

$ \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:

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

**redman223** · September 4th 2008, 10:25 PM

5y-arctan(5y)

Is this a solution to the original integral?
$ \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.

**Moo** · September 5th 2008, 04:13 AM

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.

$ \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:

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

Please don't forget dy.

$\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.

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

In the first one, calculate and then, divide by 25, properly.
In the second one, substitute carefully t=5y, because you did it wrong..

**11rdc11** · September 5th 2008, 07:29 AM

oops i forgot multiply my final answer by 1/125

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