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

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- Jun 8th 2007, 10:10 AMvietintegral
evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$

- Jun 8th 2007, 10:21 AMJhevon
- Jun 8th 2007, 10:45 AMtopsquark
Or you can use partial fractions:

$\displaystyle x^2 + 6x = x(x + 6)$

So let's find an A and B such that

$\displaystyle \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$

Well...

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

Comparing the two expressions gives me

A + B = 1

6A = 3

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

So

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

Can you finish this?

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

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

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

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

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

is this right sofar? - Jun 8th 2007, 10:52 AMJhevon
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

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

Let $\displaystyle u = x^2 + 6x$

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

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

So our integral becomes:

$\displaystyle \frac {1}{2} \int \frac {1}{u}du$ - Jun 8th 2007, 10:54 AMJhevon
- Jun 8th 2007, 11:02 AMviet
yeah i understand them, needed more practice on it. thanks for the help topsquark & Jhevon :)

- Jun 8th 2007, 11:52 AMKrizalid
$\displaystyle \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$