Partial Fraction Decomposition Evaluation

[shamieh]

Ok I'm stuck

I have \(\displaystyle \int \frac{x^2 - 5x + 16}{(2x + 1)(x - 2)^2} \, dx\)

and I got to this part:

\(\displaystyle x^2 - 5x + 16 = A(x - 2)^2 + B(x - 2)(x + \frac{1}{2}) + c(x + \frac{1}{2})\)So do i need to distribute all of these and factor out or is there a simpler way? I found a solution where they are just saying oh x = 2 and plugging it in but I am so lost on how they are getting that random value!

http://www.slader.com/textbook/9780538497909-stewart-calculus-early-transcendentals-7th-edition/493/exercises/20/#

 

[shamieh]

Ok, I see how they got x = 2. they distributed the a(x-2)^2 and then factored it out to get x = 2. Now I'm lost on how they get 5/2c = 10...Where does the 5/2 come from ?

 

[Guest2]

You're missing that if you have an identity any value that you plug into the identity returns a true value. Remember how $(x+1)(x+2) = x^2+3x+2$. How about if I hid the 3 and replaced it with $b$ instead, and told you that $(x+1)(x+2) = x^2+bx+2$. You can plug any number in this equality and get $b$ / true identity. For example, $x = 1$ gives $6 = 1+b+2$ so $3 = b$. In your case they plugged in $x = 2$ because it makes barackets containing A and B zero, leaving out the one containing C.

$\displaystyle \underbrace{(2)^2 - 5(2) + 16}_{10} = A\underbrace{(2 - 2)^2}_{0} + B\underbrace{(2 - 2)}_{0}(x + \frac{1}{2}) + c\underbrace{(2 + \frac{1}{2})}_{5/2}$

Get it?

 

[soroban]

Hello, shamieh!

$$\int \frac{x^2 - 5x + 16}{(2x + 1)(x - 2)^2} \, dx$$

Set-up: $$\frac{x^2-5x+16}{(2x+)(x-2)^2} \:=\:\frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$$

$$x^2-5x+16 \:=\:A(x-2)^2 + B(2x+1)(x-2) + C(2x+1)$$

This statement is an identity.. It is true for any value of $$x.$$


Let $$x=2\!:\;2^2-5(2)+16 \:=\:A(0^2) + B(5)(0) + C(5)$$

. . which simplifies to: .$$10 \,=\,5C \;\;\;\Rightarrow\;\;\; \boxed{C = 2}$$Let $$x = 0\!:\;0^2-5(0)+16 \:=\:A(4) + B(1)(-2) + C(1)$$

. . which simplifies to: .$$16 \,=\,4A-2B + 2 \;\;\;\Rightarrow\;\;\;4A - 2B \,=\,14$$Let $$x = 1\!:\;1^2-5(1)+16 \:=\:A(-1)^2 + B(3)(-1) + C(3)$$

. . which simplifies to: .$$12 \,=\,A-3B+6 \;\;\;\Rightarrow\;\;\; A - 3B\,=\,6$$Solve the system: .$$\boxed{A = 3},\;\boxed{B = \text{-}1}$$Therefore:

. .$$\frac{x^2-5x+16}{(2x+1)(x-2)^2} \;=\;\frac{3}{2x+1} - \frac{1}{x-2} + \frac{2}{(x-2)^2}$$

 

[BestMethod]

\(\displaystyle (x^2-5x+16)/(2x+1)(x-2)^2=a/(2x+1) + b/(x-2) + c/(x-2)^2\)
To find a,multiply both sides by \(\displaystyle (2x+1)\);

\(\displaystyle (x^2-5x+16)/(x-2)^2=a + b(2x+1)/(x-2) + c(2x+1)/(x-2)^2\)

Now,set 2x+1=0 or x=-1/2,

3=a+0+0=a

To find c,multiply both sides by \(\displaystyle (x-2)^2\);

\(\displaystyle (x^2-5x+16)/(2x+1)=a(x-2)^2/(2x+1) + b(x-2) + c\)

Now,set (x-2)=0 or x=2,

2=0+0+c=c

Now we know the values of a&c.So,to find b,set x=any number but -1/2 & 2,say 0,

4=3-b/2+1/2--> b=-1

So,the original fraction=\(\displaystyle 3/(2x+1) -1/(x-2) + 2/(x-2)^2\)

and the integrating the new fractions is easy.