Question

How do you use partial fraction decomposition to decompose the fraction to integrate `(-x - 38)/(2x^2 + 9x - 5)`?

Answer 1

`int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c`

Explanation:

First we need to factorise the denominator.

To do this we need to find factors of `-10` (coefficient of `x^2 xx` coefficient of constant`= 2 xx -5`) that add up tp 9 (the coefficient of `x=9`). So the factors we seek are `-1` and `10`

`:. 2x^2+9x-5 = 2x^2 +10x - x -5`
`:. 2x^2+9x-5 =2x(x+5) - (x+5)`
`:. 2x^2+9x-5 =(x+5)(2x-1)`

Therefore we can write the integrand as follows:

`(-x-38)/(2x^2+9x-5) = (-x-38)/((x+5)(2x-1))`

And thge partial fraction decomposition will be:

`\ \ \ \ \ (-x-38)/(2x^2+9x-5) = A/(x+5) + B/(2x-1)`
`:. (-x-38)/(2x^2+9x-5) = (A(2x-1)+B(x+5))/((x+5)(2x-1))`
`:. \ \ \ \ (-x-38) = A(2x-1)+B(x+5)`

Subs `x=-5 => -(-5)-38=A(-10-1) + 0`
`:. -11A=-33 => A=3`

And Sub `x=1/2=>-1/2-38 = B(1/2+5)`
`:. 11/2B=-77/2 => B=-7`

Hence, the partial fraction decomposition of the integrand is:

`(-x-38)/(2x^2+9x-5) = 3/(x+5) -7/(2x-1)`

And so;

`int(-x-38)/(2x^2+9x-5) dx= int 3/(x+5) -7/(2x-1) dx`

And integrating we get:

`int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c`