partial fractions: (6x+7)/(x+2)^2

[thebenji]

I have to solve the indefinite integral of:

(6x+7)/(x+2)^2

So I split it:

(6x+7)/(x+2)^2 = A/(x+2) + B/(x+2)^2

Therefore

6x+7 = A(x+2)^2 + B(x+2)

How do I find A and B? Am I even on the right track? Thanks.

 

[skeeter]

(6x+7)/(x+2)^2 = A/(x+2) + B/(x+2)^2

correction ...

6x+7 = A(x+2) + B

6x+7 = Ax + 2A + B

A has to = 6

2A + B has to equal 7 ... B = -5

 

[thebenji]

Thank you, I can see why the B does not need an (x+2) and the A only needs an (x+2) to the first power.

 

[soroban]

Hello, thebenji!

I have to solve: \(\displaystyle \L\:\int\frac{6x\,+\,7}{(x\,+\,2)^2}\,dx\)

So I split it: $\displaystyle \:\frac{6x\,+\,7}{(x\,+\,2)^2} \:=\:\frac{A}{x\,+\,2}\,+\,\frac{B}{(x\,+\,2)^2}$

Therefore: $\displaystyle \:6x\,+\,7 \:= \:A(x\,+\,2)^2\,+\,B(x\,+\,2)$ . . . no

How do I find $\displaystyle A$ and $\displaystyle B$?

You had: \(\displaystyle \L\:\frac{6x\,+\,7}{(x\,+\,2)^2} \:=\:\frac{A}{x\,+\,2}\,+\,\frac{B}{(x\,+\,2)^2}\)


Multiply through by the LCD, $\displaystyle (x+2)^2$

. . and we have: \(\displaystyle \L\:6x\,+\,7\;=\;A(x\,+\,2)\,+\,B\)



Now substitute some "good" values for $\displaystyle x$.

Let \(\displaystyle x\,=\,-2:\;\;6(-2)\,+\,7\;=\;A(-2\,+\,2) \,+\,B\;\;\Rightarrow\;\;\L B\,=\,-5\)

Let $\displaystyle x\,=\,0:\;\;6(0)\,+\,7\;=\;A(0\,+\,2)\,+\,B\;\;\Rightarrow\;\;7 \:=\:2A\,+\,B$
. . . . . . . . . Since $\displaystyle B = -5$, we have: \(\displaystyle \,7 \:=\:2A\,-\,5\;\;\Rightarrow\;\;\L A\,=\,6\)


Therefore: \(\displaystyle \L\:\frac{6x\,+\,7}{(x\,+\,2)^2} \;= \;\frac{6}{x\,+\,2}\,-\,\frac{5}{(x\,+\,2)^2}\)