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))
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))
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.
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!