Integration by Partial Fractions

[tangibleLime]

Homework Statement

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

Homework Equations

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.

 

[jbunniii]

It looks OK to me. Who told you the answer was wrong?

 

[jbunniii]

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

 

[tangibleLime]

Got it... equivalent to the second option.

 

[jbunniii]

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

$$6 \ln(3) = 6 \ln(2(3/2)) = 6\ln(2) + 6\ln(3/2)$$