Question

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

Answer 1

`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`