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

Question

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

Isaiah Allen · Accepted Answer

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

A marginally distinct method:

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#

Divide the percentage:

#=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 provides us with

#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#.

Finish the square in the denominator for the subsequent integral.

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

This should be similar to 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#

Putting everything together,

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

When we combine this with the earlier-derived expression, we find 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#

Adam Adkins · Answer

undefined The integral of $ \frac{2x}{x^2+6x+13} $ with respect to $ x $ is $ \ln|x^2+6x+13| + C $, where $ C $ is the constant of integration.