How do you use partial fraction decomposition to decompose the fraction to integrate (7)/(x^2+13x+40)7x2+13x+40?

1 Answer
Mar 7, 2018

The integral equals 7/3ln|(x + 5)/(x +8)| + C73lnx+5x+8+C

Explanation:

We wish to find factors in the denominator. The trick is to find two numbers that multiply to 4040 and add to 1313. Clearly these will be 88 and 55.

I = int 7/((x+ 5)(x+ 8))dxI=7(x+5)(x+8)dx

Now we can decompose in partial fractions.

A/(x+ 5) + B/(x +8) = 7/((x +5)(x + 8))Ax+5+Bx+8=7(x+5)(x+8)

A(x + 8) + B(x + 5) = 7A(x+8)+B(x+5)=7

Ax + 8A + Bx + 5B = 7Ax+8A+Bx+5B=7

(A + B)x + (8A + 5B) = 7(A+B)x+(8A+5B)=7

Now we have a system of equations.

{(A + B = 0), (8A + 5B = 7):}

Substituting the first equation into the second we see that

8A + 5(-A) = 7

3A = 7

A = 7/3

Now clearly B = -7/3 because A+ B =0. The integral becomes:

I = int7/(3(x + 5)) - 7/(3(x + 8)) dx

I= 7/3ln|x +5| - 7/3ln|x + 8| + C

I = 7/3ln|(x + 5)/(x +8)| + C

Hopefully this helps!