Integral Help!

• May 19th 2008, 10:21 AM
jschlarb
Integral Help!
I have this problem (forgive me, i haven't gotten the math controls down yet):

Integral 7x/(1+3x^2)^2 dx

here's what I've done so far:
A/(1+3x^2) + B/(1+3x^2)^2
= A(1+3x^2)+B(1+3x^2)/ (1+3x^2)^2
= A(1+3x^2)+B

I'm stuck right there. I can't figure out how to solve for A and B. Any help is greatly appreciated.

Cheers
• May 19th 2008, 10:34 AM
galactus
No need for partial fractions.

$7\int\frac{x}{1+3x^{2}}dx$

Let $u=1+3x^{2}, \;\ du=6xdx, \;\ \frac{1}{6}du=xdx$

$\frac{7}{6}\int\frac{1}{u}du$

$\frac{7}{6}ln(u)$

$\frac{7}{6}ln(3x^{2}+1)$
• May 19th 2008, 10:38 AM
flyingsquirrel
Hello

Quote:

A/(1+3x^2) + B/(1+3x^2)^2
The partial fractions decomposition of $\frac{7x}{1+3x^2}$ is $\frac{\alpha}{x-a}+\frac{\beta}{x-b}$ where $a$ and $b$ are the two complex roots of $1+3x^2$
• May 19th 2008, 10:50 AM
jschlarb
Quote:

Originally Posted by galactus
No need for partial fractions.

$7\int\frac{x}{1+3x^{2}}dx$

Let $u=1+3x^{2}, \;\ du=6xdx, \;\ \frac{1}{6}du=xdx$

$\frac{7}{6}\int\frac{1}{u}du$

$\frac{7}{6}ln(u)$

$\frac{7}{6}ln(3x^{2}+1)$

The denominator is supposed to be squared

$7\int\frac{x}{{(1+3x^{2})}^2}dx$

Does that change anything?
• May 19th 2008, 11:29 AM
Moo
Hello,

Quote:

Originally Posted by jschlarb
The denominator is supposed to be squared

$7\int\frac{x}{{(1+3x^{2})}^2}dx$

Does that change anything?

You can directly calculate the integral, by substituting $u=1+3x^2$
--> $du=6x dx \implies dx=\frac{du}{6x}$

$7 \int \frac{x}{(1+3x^2)^2} dx=7 \int \frac{x}{u^2} \cdot \frac{1}{6x} dx=\frac 76 \int \frac{du}{u^2}=-\frac 76 \cdot \frac 1u$

:)
• May 19th 2008, 11:32 AM
Moo
If you want to calculate it by partial fractions :

You're looking for A, B, C and D such that :

$\frac{x}{(1+3x^2)^2}=\frac{Ax+B}{1+3x^2}+\frac{Cx+ D}{(1+3x^2)^2}$, which is, I think you agree with it, quite complicated :D

You can look at this link : Quick Guide Partial Fractions :)