# integral

• Jun 8th 2007, 10:10 AM
viet
integral
evalute: $\int \frac{x + 3}{x^2 + 6 x}dx$
• Jun 8th 2007, 10:21 AM
Jhevon
Quote:

Originally Posted by viet
evalute: $\int \frac{x + 3}{x^2 + 6 x}dx$

use substitution. u = x^2 + 6x
• Jun 8th 2007, 10:45 AM
topsquark
Quote:

Originally Posted by viet
evalute: $\int \frac{x + 3}{x^2 + 6 x}dx$

Or you can use partial fractions:
$x^2 + 6x = x(x + 6)$

So let's find an A and B such that
$\frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$

Well...
$\frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}$

Comparing the two expressions gives me
A + B = 1
6A = 3

Giving me that $A = B = \frac{1}{2}$.

So
$\int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}$

Can you finish this?

-Dan
• Jun 8th 2007, 10:49 AM
viet
heres what i've done:

$\int \frac{x + 3}{x^2 + 6 x}dx =$

$\int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx$

$u = x^2+6x, du = 2x+6$

$ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx$

is this right sofar?
• Jun 8th 2007, 10:52 AM
Jhevon
Quote:

Originally Posted by topsquark
Or you can use partial fractions:
$x^2 + 6x = x(x + 6)$

So let's find an A and B such that
$\frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$

Well...
$\frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}$

Comparing the two expressions gives me
A + B = 1
6A = 3

Giving me that $A = B = \frac{1}{2}$.

So
$\int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}$

Can you finish this?

-Dan

That's nice! never thought of doing partial fractions. substitution was the way i immediately saw to do it. and since it's one of the easiest integration techniques, i didn't bother looking for another

$\int \frac {x + 3}{x^2 + 6x}dx$

Let $u = x^2 + 6x$

$\Rightarrow du = 2x + 6dx = 2(x + 3) dx$

$\Rightarrow \frac {1}{2}du = (x + 3)dx$

So our integral becomes:

$\frac {1}{2} \int \frac {1}{u}du$
• Jun 8th 2007, 10:54 AM
Jhevon
Quote:

Originally Posted by viet
heres what i've done:

$\int \frac{x + 3}{x^2 + 6 x}dx =$

$\int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx$

$u = x^2+6x, du = 2x+6$

$ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx$

is this right sofar?

dont think so. how did you get $\ln ( x + 6)$?

you didnt split the integral incorrectly, but i think you're getting yourself into more trouble than it's worth. use topsquark's or my method. do you understand them?
• Jun 8th 2007, 11:02 AM
viet
yeah i understand them, needed more practice on it. thanks for the help topsquark & Jhevon :)
• Jun 8th 2007, 11:52 AM
Krizalid
$\int {\frac{{x + 3}}
{{x^2 + 6x}}~dx} = \frac{1}
{2}\int {\frac{{\left( {x^2 + 6x} \right)^\prime }}
{{x^2 + 6x}}~dx} = \frac{1}
{2}\ln (x^2 + 6x) + k$