[SOLVED] Integration by parts, evil arctan

redman223 · #1 $Old$ 09-04-2008, 08:45 PM

$\int x~arctan(7x)~dx$

u= arctan(7x)

du= $\frac {1}{1+49x^2}$

dv= x~dx

v= $\frac {x^2}{2}$

This is where I am at so far:
$x^2arctan(7x) - \frac {1}{2}\int \frac {x^2}{1+49x^2}$

Chop Suey · #2 $Old$ 09-04-2008, 08:56 PM

Use polynomial long division for the second integral. You will get 2 simpler integrals.

Chop Suey · #3 $Old$ 09-04-2008, 09:01 PM

Or, if you want to do things quickly, pull 49 as a factor:
$\frac{x^2}{49(\frac{1}{49} + x^2)}$

Can you see how to simplify this quickly without resorting to long division? Hint: Add $\frac{1}{49}$ and subtract $\frac{1}{49}$ in the numerator.

redman223 · #4 $Old$ 09-05-2008, 12:02 AM

We have only talked about the long division method, but I need a little help with the 2 other integrals. What will the other two integrals be?

So I see that after long division, I get 1/49, but unfortunately I don't really know what to do next.

$\int \frac {1}{49} (1- \frac {x^2}{1+49x^2})$

Is this correct? What is next?

Chop Suey · #5 $Old$ 09-05-2008, 12:09 AM

Quote:

Originally Posted by redman223 $View Post$

$\int \frac {1}{49} (1- \frac {\color{red}x^2}{1+49x^2})$

That $x^2$ shouldn't be there. It's supposed to be:

$\frac{1}{1+49x^2}$

Once you figure that out, notice that:

$\frac{1}{1+49x^2} = \frac{1}{1+(7x)^2}$

redman223 · #6 $Old$ 09-05-2008, 12:31 AM

$\int \frac {1}{49} (1- \frac {1}{1+49x^2})$

So what happens to the first part of the equation? With regards to the 1/49 and the 1-?

Chop Suey · #7 $Old$ 09-05-2008, 12:47 AM

$\int \frac {1}{49} (1- \frac {1}{1+49x^2})$ can be split into:

$= \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+49x^2}$

$= \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+(7x)^2}$

Matt Westwood · #8 $Old$ 09-05-2008, 12:49 AM

it's $\int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx$ okay? So you can split it into two:

$\int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx = \int \frac 1 {49}\left({1}\right) dx - \int \frac 1 {49}\left({\frac {1}{1 + 49x}}\right) dx$

redman223 · #9 $Old$ 09-05-2008, 01:38 AM

Thanks for clarifying that up for me. Now I can see a solution using u substitution.

redman223 · #10 $Old$ 09-05-2008, 02:06 AM

Well I thought I did, but I guess not. I am stuck once again.

$-\frac {1}{2}\int \frac {1}{49}~dx = -\frac {1}{2}(\frac {1}{49}x)$

If thats the case, then
$-\frac {1}{2}(\frac {1}{49}x - \frac {1}{49}\int \frac {1}{1+49x^2}~dx)$

u= 7x
du= 7

So in order to get the 7 in there I have to multiply 1/49 by 1/7 in order to get 1/343 right? Then would the answer just be:

$-\frac {1}{2}(\frac {arctan(7x)}{343})$

for the last part of the equation?