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

Explanation

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