Question

What is `int (2x)/(x^2+6x+13) dx`?

Answer 1

`ln(x^2+6x+13)-3arctan((x+3)/2)+C`

Explanation:

A slightly different approach:

Since the derivative of the denominator is `2x+6`, make the following modification to the numerator:

`=int(2x+6-6)/(x^2+6x+13)dx`

Split the fraction:

`=int(2x+6)/(x^2+6x+13)dx-int6/(x^2+6x+13)dx`

For the first integral, let `u=x^2+6x+13` and `du=(2x+6)dx`.

This gives us

`int1/udu-6int1/(x^2+6x+13)dx`

`=lnabsu-6int1/(x^2+6x+13)dx`

`=ln(x^2+6x+13)-6int1/(x^2+6x+13)dx`

Note that the absolute value signs are no longer necessary since `x^2+6x+13>0` for all values of `x`.

For the next integral, complete the square in the denominator.

`ln(x^2+6x+13)-6int1/((x+3)^2+4)dx`

We will want to make this resemble the arctangent integral:

`int1/(u^2+1)du=arctan(u)+C`

Focusing on just `int1/((x+3)^2+4)dx`, we divide everything by `4`.

`=int(1/4)/((x+3)^2/4+4/4)dx=1/4int1/((x+3)^2/4+1)dx`

To make this resemble `int1/(u^2+1)du`, we set `u=(x+3)/2`.

`=1/4int1/(u^2+1)dx`

To achieve a `du` value, first note that `du=1/2dx`. Multiply the interior of the integral by `1/2` and the exterior by `2`.

`=1/2int(1/2)/(u^2+1)dx=1/2int1/(u^2+1)du=1/2arctan(u)+C`

Pulling this together,

`int1/((x+3)^2+4)dx=1/2arctan((x+3)/2)+C`

Combining this with the expression we came up with earlier, we see that the original integral equals

`=ln(x^2+6x+13)-6int1/(x^2+6x+13)dx`

`=ln(x^2+6x+13)-6(1/2)arctan((x+3)/2)+C`

`=ln(x^2+6x+13)-3arctan((x+3)/2)+C`