What is the integral of #(2x)/(x^2 + 6x + 9)# using partial fractions or #u# substitution?

There are 2 ways of performing this integral.

Let's do it by partial fractions

#(2x)/(x^2+6x+9)=(2x)/(x+3)^2#

#=A/(x+3)^2+B/(x+3)#

#=(A+B(x+3))/(x+3)^2#

The denominators are the same, we compare the numerators

#2x=A+B(x+3)#

Let #x=-3#, #=>#, #-6=A#

Coefficients of #x#

#2=B#

Therefore,

#(2x)/(x^2+6x+9)=-6/(x+3)^2+2/(x+3)#

#int(2xdx)/(x^2+6x+9)=int(-6dx)/(x+3)^2+int(2dx)/(x+3)#

#=6/(x+3)+2ln(|x+3|)+C#

You can also, do

#u=x+3#, #=>#, #dx=du#

#x=u-3#

#int(2xdx)/(x+3)^2=int((2)(u-3)du)/(u^2)#

#=2int(1/u-3/u^2)du#

#=2lnu+6/u#

#=2ln(|x+3|)+6/(x+3)+C#

Hope that this is helpful

To be honest, I was on the same boat as you until you mentioned integration by partial fractions. That's when it clicked for me. If you had not mentioned partial fractions, I probably would not know how to have solved it. Here goes:

For partial fractions, you would first factor the denominator:

#x^2+6x+9=(x+3)^2#

Now you set it with variables A and B so that:

#A/(x+3) + B/(x+3)^2 = (2x)/(x+3)^2# (from integral)

(Remember to account for the multiplicity of the factor.)
Multiply everything by #(x+3)^2# to get:

#A(x+3) + B = 2x#

Let's distribute.

#Ax +(Axx3) + B = 2x#

This may lead somewhere. Let's set #x=0# to get rid of a few terms:

#(Axx3) + B=0#

#3A=-B#

#A=-B/3#

Plug this in so that #int A/(x+3)dx +int B/(x+3)^2dx#, and manipulate it to equal:

#int (-B"/"3)/(x+3)dx +int B/(x+3)^2dx#

#-1/3int B/(x+3)dx +int B/(x+3)^2dx#

This would equal #-B/3 ln |x+3| - B/(x+3)# due to integration properties, which equals #int(2x)/(x^2+6x+9)dx#

To solve for B, set the equation as such:

#-B/3 ln |x+3| - B/(x+3) =int(2x)/(x^2+6x+9)dx#

Factor B:

#B(-1/3ln |x+3| - 1/(x+3))=int(2x)/(x^2+6x+9)dx#

Divide by #(-1/3ln |x+3| - 1/(x+3))# on both sides:

#B=(int(2x)/(x^2+6x+9)dx)/(-1/3ln |x+3| - 1/(x+3))#

Input this as B to get:

#-1/3((int(2x)/(x^2+6x+9)dx)/(-1/3ln |x+3| - 1/(x+3))) ln |x+3| - ((int(2x)/(x^2+6x+9)dx)/(-1/3ln |x+3| - 1/(x+3))) 1/(x+3)#

Phew! This is your final result, after you divide out the starting integral.

Let's do it your way (sort of):

#int (2x)/(x^2+6x+9) dx = #

Add 0 to the numerator in the form of +6-6:

#int (2x+6-6)/(x^2+6x+9) dx = #

We can break this into two fractions:

#int (2x+6)/(x^2+6x+9)-6/(x^2+6x+9) dx = #

We can make this two integrals:

#int (2x+6)/(x^2+6x+9) dx -6int1/(x^2+6x+9) dx = #

You know how to do the first integral:

let #u = x^2+6x+9#, then #du = 2x+6dx#:

#int 1/u du -6int1/(x^2+6x+9) dx = #

#ln|u| -6int1/(x^2+6x+9) dx = #

Reverse the u substitution:

#ln|x^2+6x+9| -6int1/(x^2+6x+9) dx = #

Let's factor the denominator in the integral:

#ln|x^2+6x+9| -6int1/(x+3)^2 dx = #

Let #u = x+3#, then #du = dx#:

#ln|x^2+6x+9| -6int1/u^2 du = #

Write as a negative exponent:

#ln|x^2+6x+9| -6intu^-2 dx = #

Use the power rule to integrate:

#ln|x^2+6x+9| +6u^-1+C = #

Reverse the u substitution:

#ln|x^2+6x+9| +6/(x+3)+C larr # answer.

As:

#x^2+6x+9 = (x+3)^2#

you can also integrate by parts:

#int (2x)/(x+3)^-2 dx = int 2x d(-1/(x+3))#

#int (2x)(x+3)^-2 dx = -(2x)/(x+3) + 2int dx/(x+3)#

#int (2x)(x+3)^-2 dx = -(2x)/(x+3) + 2ln abs(x+3)+C#