definite integral with exponential

writing numerator as (e^ax+1) - (e^bx+1) then dividing the integral into two parts ,

I = ₀∫^∞ dxx(e^bx+1) - ₀∫^∞ dxx(e^ax+1)

now let f(a) = ₀∫^∞ dxx(e^ax+1)

f '(a)= ₀∫^∞ - e^ax .x .dxx(e^ax+1)² = 1a₀∫^∞ a.e^axdx(e^ax+1)²

f '(a) = -1a

f(a) = - ln (a) +c

now I = f(b) - f(a) = ln ab

Very Nice Aditiya.

Same method used by you in Solving \hspace{-16}\mathbf{\int\frac{1}{(x^2+a^2)^2}dx} in goiit.com