MHB Partial Fraction Decomposition Help - Calculus BC

[MarkFL]

Here is the question:

Help with Calculus BC: partial fractions!?

Thank you for your help guys. I wrote the problem wrong in the previous question :(

integral (4x^2+x-2)/(x^3+2x^2-3x) dx

howwwww?

Here is a link to the question:

Help with Calculus BC: partial fractions!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello living higher:

We are given to decompose:

$\displaystyle \frac{4x^2+x-2}{x^3+2x^2-3x}$

The first step is to factorize the denominator:

$x^3+2x^2-3x=x(x^2+2x-3)=x(x-1)(x+3)$

Now we assume the decomposition will take the form:

$\displaystyle \frac{4x^2+x-2}{x(x-1)(x+3)}=\frac{A}{x+3}+\frac{B}{x}+\frac{C}{x-1}$

Now, rather than set up a linear system of equations, I am going to use a shortcut method called the Heaviside cover-up method. Look at the first term on the right. The root of the denominator is $x=-3$.

To find the value of $A$, we "cover-up" the factor $x+3$ to get:

$\displaystyle \frac{4x^2+x-2}{x(x-1)}$

and we evaluate this at $x=-3$ to find:

$\displaystyle A=\frac{4(-3)^2+(-3)-2}{(-3)((-3)-1)}=\frac{31}{12}$

Next we look at the second term in the decomposition and we see the root of the denominator is $x=0$, and covering up the factor $x$ on the left, and evaluating it for $x=0$, we find:

$\displaystyle B=\frac{4(0)^2+(0)-2}{((0)-1)((0)+3)}=\frac{2}{3}$

And finally, we look at the third term in the decomposition and we see the root of the denominator is $x=1$, and covering the the factor $x-1$ on the left and evaluation it for $x=1$, we find:

$\displaystyle C=\frac{4(1)^2+(1)-2}{(1)((1)+3)}=\frac{3}{4}$

And now we may state:

$\displaystyle \frac{4x^2+x-2}{x^3+2x^2-3x}=\frac{31}{12(x+3)}+\frac{2}{3x}+\frac{3}{4(x-1)}$

Now we may directly integrate:

$\displaystyle \int\frac{31}{12(x+3)}+\frac{2}{3x}+\frac{3}{4(x-1)}\,dx=\frac{31}{12}\ln|x+3|+\frac{2}{3}\ln|x|+ \frac{3}{4}\ln|x-1|+C$