# Integration by Partial Fractions

tangibleLime

## Homework Statement

$$\int_0^1 \! \frac{x-4}{x^2+3x+2} dx$$

## The Attempt at a Solution

Factoring out the denominator...

$$x^2+3x+2 = (x+1)(x+2)$$​

Breaking up the main fraction into the separate fractions. Since both are linear, only a single constant is needed on top.

$$\frac{A}{x+1}+\frac{B}{x+2}$$​

Fining the coefficients (A and B) by choosing x values that make the other coefficient go to zero.

$$x-4 = A(x+2) + B(x+1)$$

$$x=-2$$
$$-6=B(-1)$$
$$6=B$$

$$x=-1$$
$$-5=A(1)$$
$$-5=A$$​

Back in the integral they go. Also, breaking up the main integral into a sum of integrals.

$$\int \frac{-5}{x+1}dx + \int \frac{6}{x+2} dx$$

$$-5ln(x+1) + 6ln(x+2) + c$$

Now using the fundamental theorem of calculus to find the value.

$$(-5ln(1+1) + 6ln(1+2))-(-5ln(0+1) + 6ln(0+2))$$​

Result:

$$6ln(3)-11ln(2)$$​

Wrong. Aw.

Homework Helper
Gold Member
It looks OK to me. Who told you the answer was wrong?

Homework Helper
Gold Member
P.S. Wolfram Alpha's answer agrees numerically with yours.

tangibleLime
The online homework system... though it gave me multiple options, and none of them were this. I suppose I could try to figure out which is equal, if my work is indeed correct.

http://mikederoche.com/temp/options.png [Broken]

Last edited by a moderator:
tangibleLime
Got it... equivalent to the second option.

It looks like they really want you to have a $\ln(3/2)$ term for some reason. OK then:
$$6 \ln(3) = 6 \ln(2(3/2)) = 6\ln(2) + 6\ln(3/2)$$