Question

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

Answer 1

The answer is `=6/(x+3)+2ln(|x+3|)+C`

Explanation:

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

Answer 2

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

Explanation:

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.

Answer 3

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.

Answer 4

`int (2x)/(x^2+6x+9) dx = -(2x)/(x+3) + 2ln abs(x+3)+C`

Explanation:

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`