`f(x)=(12+8x-x^2)/((2-x)(4+x^2))`

First we need to use partial fractions to break the function in to two.

`(12+8x-x^2)/((2-x)(4+x^2))=A/(2-x)+(Bx+C)/(4+x^2)`

`12+8x-x^2=A(4+x^2)+(Bx+C)(2-x)`

`12+8x-x^2=A(4+x^2)+2Bx-Bx^2+2C-Cx`

`12+8x-x^2=x^2(A-B)+x(2B-C)+(2C+4A)`

By comparing coefficients;

`x^2rarr -1=A-B` ------------(1)

`xrarr 8=2B-C` ----------------(2)

`Constant rarr 12 =2C+4A` -------(3)

solving (1),(2) and (3) give you;

`A=3`

`B=4`

`C=0`

`12+8x-x^2/(2-x)(4+x^2) = 3/(2-x) + 4x/(4+x^2)`

`int_0^1 f(x) dx...

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`f(x)=(12+8x-x^2)/((2-x)(4+x^2))`

First we need to use partial fractions to break the function in to two.

`(12+8x-x^2)/((2-x)(4+x^2))=A/(2-x)+(Bx+C)/(4+x^2)`

`12+8x-x^2=A(4+x^2)+(Bx+C)(2-x)`

`12+8x-x^2=A(4+x^2)+2Bx-Bx^2+2C-Cx`

`12+8x-x^2=x^2(A-B)+x(2B-C)+(2C+4A)`

By comparing coefficients;

`x^2rarr -1=A-B` ------------(1)

`xrarr 8=2B-C` ----------------(2)

`Constant rarr 12 =2C+4A` -------(3)

solving (1),(2) and (3) give you;

`A=3`

`B=4`

`C=0`

`12+8x-x^2/(2-x)(4+x^2) = 3/(2-x) + 4x/(4+x^2)`

`int_0^1 f(x) dx `

`= int_0^1 3/(2-x)dx + int_0^1 (4x)/(4+x^2)dx`

`=(-3)[ln|2-x|]_0^1+2[ln|4+x^2|]_0^1`

`=(-3)(0-ln2)+2(ln5-ln4)`

`=2ln5-2ln4+3ln2`

`=2ln5-4ln2+3ln22`

`=2ln5-ln2`

`=ln5^2-ln2`

`=ln25-ln2`

`=ln(25/2)`

*So the required answer is proved.*

**Further Reading**