How do you integrate #1/((x+6)(x^2+3))# using partial fractions?

1 Answer
Nov 5, 2016

#1/78ln((x+6)^2/(x^2+3))+2/(13sqrt3)arctan(x/sqrt3)+C#

Explanation:

Decomposing the fraction:

#1/((x+6)(x^2+3))=A/(x+6)+(Bx+C)/(x^2+3)#

So:

#1=A(x^2+3)+(Bx+C)(x+6)#

#1=Ax^2+3A+Bx^2+6Bx+Cx+6C#

Sorting by #x#:

#0x^2+0x+1=x^2(A+B)+x(6B+C)+(3A+6C)#

Comparing the coefficients:

#{(A+B=0),(6B+C=0),(3A+6C=1):}#

From the first equation we see that #B=-A#. Substituting this into the second equation, we see that #-6A+C=0#. Multiplying the third equation by #2#, we see that #6A+12C=2#. Adding this to #-6A+C=0#, we get #13C=2# so #C=2/13#.

Using this value in the second equation, we see that #A=1/39#. Thus #B=-1/39#.

Using these values:

#1/((x+6)(x^2+3))=(1/39)/(x+6)+(-1/39x+2/13)/(x^2+3)#

#1/((x+6)(x^2+3))=1/39 1/(x+6)+1/39(-x+6)/(x^2+3)#

So:

#I=int1/((x+6)(x^2+3))dx=1/39int1/(x+6)dx+1/39int(-x+6)/(x^2+3)#

And:

#I=1/39int1/(x+6)dx-1/39intx/(x^2+3)dx+6/39int1/(x^2+3)dx#

Letting #u=x+6# for the first integral, so that #du=dx#:

#I=1/39int1/udu-1/39intx/(x^2+3)dx+2/13int1/(x^2+3)dx#

#I=1/39lnabsu-1/39intx/(x^2+3)dx+2/13int1/(x^2+3)dx#

For the next integral, let #v=x^2+3# so #du=2xdx#:

#I=1/39lnabsu-1/78int(2x)/(x^2+3)dx+2/13int1/(x^2+3)dx#

#I=1/39lnabsu-1/78int1/vdv+2/13int1/(x^2+3)dx#

#I=1/39lnabsu-1/78lnabsv+2/13int1/(x^2+3)dx#

Before moving onto the next integral, we can do a very sneaky simplification:

#I=2/78lnabsu-1/78lnabsv+2/13int1/(x^2+3)dx#

#I=1/78lnabs(u^2)-1/78lnabsv+2/13int1/(x^2+3)dx#

#I=1/78lnabs(u^2/v)+2/13int1/(x^2+3)dx#

With #u=x+6# and #v=x^2+3#:

#I=1/78lnabs((x+6)^2/(x^2+3))+2/13int1/(x^2+3)dx#

Note the absolute value bars aren't necessary since the function inside the logarithm is always positive:

#I=1/78ln((x+6)^2/(x^2+3))+2/13int1/(x^2+3)dx#

For the remaining integral, there are two courses of action. The first would be to use the arctangent integral formula: #int1/(u^2+a^2)du=1/aarctan(u/a)+C#.

The other, which I prefer since you don't have to remember the formula, is to use trigonometric substitution.

So, for #int1/(x^2+3)dx#, let #x=sqrt3tantheta#. Thus #dx=sqrt3sec^2thetad theta# and:

#I=1/78ln((x+6)^2/(x^2+3))+2/13int1/(3tan^2theta+3)(sqrt3sec^2thetad theta)#

#I=1/78ln((x+6)^2/(x^2+3))+2/(13sqrt3)intsec^2theta/(tan^2theta+1)d theta#

Since #sec^2theta=1+tan^2theta#:

#I=1/78ln((x+6)^2/(x^2+3))+2/(13sqrt3)intd theta#

From #x=sqrt3tantheta# we see #theta=arctan(x/sqrt3)#.

#I=1/78ln((x+6)^2/(x^2+3))+2/(13sqrt3)arctan(x/sqrt3)+C#