ज्ञात करें `int (dx)/(x(x^n +1))`

Explanation

माना की `z=x^n +1` तो `dz = nx^(n-1) dx`
अब, `I =int (dx)/(x(x^n+1)) = int (dz)/(nx^(n-1)*x (x^(n)+1))`
`=1/n int(dx)/(x^n (x^n+1))=1/n int (dz)/((z-1)z)" ...(1)"`
माना की ` 1/(z(z-1))=A/z+B(z-1) =(A(z-1)+Bz)/(z(z-1)" ...(2)"`
`:." "A(z-1)+Bz=1" ...(3)"`
`z=1` रखे पर, `B=1`
`:." " (1)` से, `I=I/n int (dz)/(z(z-1))=1n""int(-1/z+1/(z-1))dz`
`=1/n [-log |z|+log |z-1|]+c`
`=1/nm log|(z-1)/z|+c =1/n "log"|x^n/(x^n+1)|+c`