Let `f(x) = (12 + 8x-x^2) /((2-x)(4 + x^2))` Show that `int_0^1f(x)dx = ln(25/2)`
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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.
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