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int(1)^(3) (dx)/(x^(2)(x+1))=(2)/(3)+log...

`int_(1)^(3) (dx)/(x^(2)(x+1))=(2)/(3)+log .(2)/(3)`

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`int_(1)^(3) (1)/(x^(2)(x+1))dx=(2)/(3)+log .(2)/(3)`
`" Let " (1)/(x^(2)(x+1))=(A)/(x)+(B)/(x^(2))+(C )/(x+1)" "(1)`
`rArr 1=A x(x+1) +B(x+1) +Cx^(2)" "......(2)`
x =0 then 1=0 +B+0`rArr` B
x=-1 then 1=0 +0+C `rArr` C=
Equating coefficients of `x^(2)`
`0=A+C rArr A=-C =-1`
`:. int_(1)^(3)(1)/(x^(2)(x+1))dx`
`=int_(1)^(3) (-(1)/(x)+(1)/(x^(2))+(1)/(x+1))dx`
`=[-log|x|-(1)/(x)+log|x+1|]_(1)^(3)`
`=[log|(x+1)/(x)|-(1)/(x)]_(1)^(3)`
`=(log |(4)/(3)|-(1)/(3))-(log|(2)/(1)|-(1)/(1))`
`=(log .(4)/(3) -log 2)+(1-(1)/(3))`
=log ((4)/(2)xx(1)/(2))+(2)/(3)=log.(2)/(3)+(2)/(3)`
Hence proved.
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