Question

Can some one help me do this integral using partial fractions (I know its simple but I'm just learning partial fractions)

Integral of (x+2)/(x^2+4x)

Answer 1

substitute: u=x^2+4x
then du=2x+4=2(x+2)

Answer 2

I know its a simple u-sub problem. But i would like to see it done with partial fractions

Answer 3

then you'll have
integral 1/2 du/u
right?
can you finish it?

Answer 4

inik... I know how to do the integral.. But i would like to see it done with partial fractions... And i even KNOW how to do partial fractions but for B I keep getting 3/2's and idk what to do with a 3/2's exactly

Answer 5

\[=1/2 \ln|x ^{2}+4x| + const\]

Answer 6

oups... let me try

Answer 7

A/(x+4)+B/x=(x+2)/(x^2+4x)
Ax+B(x+4)=x+2
so A and B are both 1/2

Answer 8

hope from here it is clear

Answer 9

I set it up with A / x + B / (x-4) then for my B value I got 3/2's ... I guess I just don't know what to do after I get a fraction

Answer 10

A/x + (Bx+C)/(x^2+4)

Answer 11

I got A=1/2, B=-1/2 and C=1
so,
1/(2x) +(-x/2 +1)/(x^2+1)
integrate

Answer 12

\[=\int\limits_{ }^{}dx/(2x) +\int\limits_{ }^{}(1-x/2)/(x ^{2}+4)=...\]
=\[1/2\int\limits_{ }^{}dx/x +\int\limits_{ }^{}(2-x)dx/2(x+4) =...\]
should work!

Answer 13

Tfraiz, I got you lol.

Answer 14

You have \[\int\limits \frac{x+2}{x^2+4x}dx\]
Start by factoring the x out of the bottom.
\[\int\limits \frac{x+2}{x(x+4)}dx=\int\limits(\frac{A}{x}+\frac{B}{x+4})dx\]
Getting a common denominator your NUMERATOR would be:
A(x+4)+Bx=x+2
Try plugging x=0 then:
A(0+4)+B(0)=0+2
Or 4A=2
A=(1/2)
Then plug in x=-4
A(-4+4)+B(-4)=-4+2
or -4B=-2
B=1/2
So you have:
\[\frac{1}{2}\int\limits \frac{dx}{x}+\frac{1}{2}\int\limits \frac{dx}{x+4}=\frac{1}{2}\ln|x|+\frac{1}{2}\ln|x+4|+C\]