ज्ञात करें ` int (log x)/((1+ log x)^2)dx`

Explanation

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