Question

How do you find `int (x+1)/(9x^2+6x+5)dx` using partial fractions?

Answer 1

`int(x+1)/(9x^2+6x+5)dx`

= `1/18ln(9x^2+6x+5)+1/9tan^(-1)((3x+1)/2)+c`

Explanation:

As the discriminant of `9x^2+6x+5` is `6^2-4xx9xx5=36-180=-144`, it being negative, cannot be factorized in rational factors.

Observing that differential of `9x^2+6x+5` is `18x+6`, let us split `(x+1)/(9x^2+6x+5)` as `(18x+18)/(18(9x^2+6x+5))` or

`(18x+6)/(18(9x^2+6x+5))+12/(18(9x^2+6x+5))`

Hence `int(x+1)/(9x^2+6x+5)dx`

= `int(18x+6)/(18(9x^2+6x+5))dx+int12/(18(9x^2+6x+5))dx`

= `1/18int(18x+6)/(9x^2+6x+5)dx+2/3int1/(9x^2+6x+5)dx`

Let us integrate first part and assume `u=9x^2+6x+5`, then `du=(18x+6)dx` and it can be written as `1/18int(du)/u=1/18lnu=1/18ln(9x^2+6x+5)`.

For second part `2/3int1/(9x^2+6x+5)dx`, we will use the identity `int(dx)/(x^2+a^2)=1/atan^(-1)(x/a)+c`. For this converting denominator into sum of two squares, we get `(9x^2+6x+1)+2^2`, hence

`2/3int1/(9x^2+6x+5)dx=2/27int1/((x^2+2/3x+1/9)+(2/3)^2)dx=2/27int1/((x+1/3)^2+(2/3)^2)dx`

Let us now substitute `u=x+1/3` and as `du=dx` this is now

`2/27int1/(u^2+(2/3)^2)du=2/27xx(3/2tan^(-1)((3u)/2))` or

`1/9tan^(-1)((3(x+1/3))/2)=1/9tan^(-1)((3x+1)/2)`

Hence, `int(x+1)/(9x^2+6x+5)dx`

= `1/18ln(9x^2+6x+5)+1/9tan^(-1)((3x+1)/2)+c`