# evaluate the following integral (x-2)/(x^2-3x-54)dx limits are 8,-5 please explain as you solve. I have most of my problems toward the end simplifying

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`int ^8_(-5)(x-2)/(x^2-3x-54) dx`

we know;

d(`x^2-3x-54` )/dx = 2x-3

In the question we have to make the top part as (2x-3)

`(x-2) = (2x-3)/2-1/2`

`(x^2-3x-54)`

`= (x-3/2)^2-54-9/4`

`= (x-3/2)^2-225/4`

`=(x-3/2)^2-(15/2)^2`

`= (x-3/2+15/2)(x-3/2-15/2)`

`= (x+6)(x-9)`

`1/(x^2-3x-54)`

`= 1/[(x+6)(x-9)]`

`= (1/15){1/(x-9)-1/(x+6)}`

`int ^8_-5(x-2)/(x^2-3x-54) dx`

`=int ^8_-5[(2x-3)/2-1/2]/(x^2-3x-54) dx`

`=int ^8_(-5) (1/2) (2x-3)/(x^2-3x-54) dx -int ^8_(-5) (1/2)(1/(x^2-3x-54) dx`

`=int ^8_(-5) (1/2) (2x-3)/(x^2-3x-54) dx -int ^8_(-5) (1/2)(1/15)[1/(x-9)-1/(x+6)] dx`

`= (1/2)[ln (x+6)(x-9)]^8_(-5)-1/30 [ln(x-9)/(x+6)]^8_(-5)`

**When you apply the limit there is a error because of the limit values lead the ln (x^2-3x-54)<0.**

**So i will give only the indefinite integral here. **

**`int (x-2)/(x^2-3x-54) dx =(1/2)[ln (x+6)(x-9)]-1/30 [ln[(x-9)/(x+6)]]+C` **

C is a constant.

**Sources:**