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Evaluate `int dx/(x(x^n+1))`

Answer» Putting `x^(n)=t,` We get `nx^(n-1)dx=dt.`
`therefore (nx^(n))/(x)dx=dt implies(1)/(x)dx=(1)/(nt)dt("note").`
`therefore int(dx)/(x(x^(n)+1))=int(dt)/(nt(t+1))=(1)/(n).int(dt)/(t(t+1)).`
`Let (1)/(t(t+1))=(A)/(t)+(B)/((t+1)).`
Then`,1-=A(t+1)+Bt.`
Putting `t=0`in (i) ,we get `A=1.`
Putting `t=-1` in (i), we get `B=-1.`
`therefore (1)/(t(t+1))={(1)/(t)-(1)/((t+1))}`
`=(1)/(n)[int(1)/(t)dt-int(1)/((t+1))dt]`
`=(1)/(n).{log|t|-log|t+1|}+C`
`=(1)/(n).log|(t)/(t|1)|+C`
`=(1)/(n)log|(x^(n))/(x^(n)+1)|+C.`


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