# Find `int_1^2 (5x-4)/((1-x+x^2)(2+x))dx`

### 1 Answer | Add Yours

To solve this we need to use partial fractions.

`(5x-4)/((1-x+x^2)(2+x))=A/(2+x) + (Bx+C)/(1-x+x^2)`

`5x-4 = A(1-x+x^2)+(Bx+C)(2+x)`

When x = -2

`-14 = 7A`

`A = -2`

`5x-4 = -2(1-x+x^2)+(Bx+C)(2+x)`

comparing coeffieicen of x^2;

`0 = -2+B`

`B = 2`

Comparing coefficient of contant term;

`-4 = -2+2C`

`C = -1`

`int_1^2(5x-4)/((1-x+x^2)(2+x))dx`

`= int_1^2 [-2/(2+x)+(2x-1)/(1-x+x^2)]dx`

`= -2int_1^2 (1)/(2+x)dx+int_1^2 (2x-1)/(1+x^2-x)dx`

`= -2[ln|2+x|]_1^2+[ln|1+x^2-x|]_1^2`

`= -2(ln4-ln3)+(ln3-ln1)`

`=3ln3-2ln4`

`=ln(27/16)`

*So the answer is;*

`int_1^2(5x-4)/((1-x+x^2)(2+x))dx=ln(27/16)`

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