# Evaluate the integral `int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6)`

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What kspcr111 said :-)

`int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6) dx`

sol:-

first just integrate it

`int(x^3 - 4x - 10)/(x^2 - x - 6) dx`

= `int x+1+(3x-4)/(x^2-x-6) dx`

`= int x dx+int 1 dx+int (3x-4)/(x^2-x-6) dx`

`= (x^2)/2 +x+int (3x-4)/((x+2)(x-3)) dx`

`= (x^2)/2 +x+int ((2/(x+2))+(1/(x-3))) dx `

`= (x^2)/2 +x+int (2/(x+2))+ int(1/(x-3)) dx`

`= (x^2)/2 +x+2*ln(x+2)+ln(x-3) `

Now let us apply the limits from 0 to 1, we get

`int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6) dx`

`=[(x^2)/2 +x+2*ln(x+2)+ln(x-3)]^1_0 `

`= (1/2 )+ 1+2ln(3)+ln(-2) -2ln(2)-ln(-3) `

` = 3/2 +ln(9) + ln(-2) - (ln(4)+ln(-3))`

`= 3/2 +ln(9*(-2)) - (ln(4*(-3)))`

`=(3/2)+ ln(-18) -ln(-12)`

` = (3/2)+ln(-18/-12)`

`=(3/2)+ln(3/2) `

Done :)