# Evaluate the definite integral ∫1/((x+9)(x62+4)) , x from -3 to 2

• Mar 2nd 2009, 07:37 PM
Kitizhi
Evaluate the definite integral ∫1/((x+9)(x62+4)) , x from -3 to 2
I've been stuck on this question for a while.
Can someone show me how to do this one, thanks.

Question is actually

∫1/((x+9)(x^2+4))
• Mar 2nd 2009, 07:57 PM
treetheta
Try partial subsitiuation

$\displaystyle \int{{{1}\over{\left(x+9\right)\,\left(x^2+4\right )}}}\,dx$
$\displaystyle \int{{{1}\over{\left(x+9\right)\,\left(x^2+4\right )}}}\,dx$$\displaystyle = A/(x+9) + (Bx+C)/(x^2+4)$

solve for A and B and then integrate it will be much easier.
• Mar 2nd 2009, 08:44 PM
mollymcf2009
Quote:

Originally Posted by Kitizhi
I've been stuck on this question for a while.
Can someone show me how to do this one, thanks.

Question is actually

∫1/((x+9)(x^2+4))

Partial Fractions:

$\displaystyle \int \frac{1}{(x+9)(x^2+4)} dx$

$\displaystyle \frac{A}{(x+9)} + \frac{Bx+C}{x^2+4}$

Get a common denominator:

$\displaystyle A(x^2 + 4) + (Bx+C)(x+9)$ * THis is equal to the bottom of your original integral

So,

$\displaystyle 1 = A(x^2 + 4) + (Bx+C)(x+9)$

Now FOIL the right side and collect like terms:
$\displaystyle = Ax^2 + 4A + Bx^2 + 9Bx + Cx + 9C$

$\displaystyle = (A + B)x^2 + (9B +C)x + (4A + 9C)$

Now you can solve for A B & C * Remember that since your numerator in your original integral was 1, you will only put 4A + 9C = 1. You always put your like terms equal to the coefficient of the like term in the numerator of your original integral, so in this case since there are no $\displaystyle x$ or $\displaystyle x^2$terms in the numerator of your integral, the other two will be = 0.

1 = 4A + 9C

0 = A + B

0 = 9B + C

So,

C = -9B
A = -B

then

1 = 4(-B) + 9(-9B)
1 = -4B - 81B
1 = -85B

$\displaystyle B = -\frac{1}{85}$

Then you can solve for the others.

When you get your A B & C, just plug them back into

$\displaystyle \frac{A}{(x+9)} + \frac{Bx+C}{x^2+4}$

and this is your new integral!
Whew, these are usually pretty long and time consuming! Hope this helps!
• Mar 2nd 2009, 08:51 PM
Kitizhi
Thanks alot mollymcf2009,
I think I can manage from here on