ज्ञात करें `int [ 1/( log x)-1/(( log x)^2)]dx`

Explanation

माना की `z= log _(e)x " "x = e^x " ":." "dx = e^z dz`
अब, `I = int[1/(log x)-1/((logx)^2)]dx =int (1/z-1/z^2)e^z dz`
`=int e^z [1/z +(-1/z^2)]dz=int e^z [f(z)+f'(z)]dz` जहां `f(z)=1/z`
`=e^z f(z)+c =e^z ""1/z +c=x/(log x)+c`