Indefinite IntegrationmediumFree

Integral of (g'/g − f'/f)·ln(g/f) | JEE

JEE Maths question with a full step-by-step solution.

Question
∫(g′(x)g(x)−f′(x)f(x))(ln⁡g(x)−ln⁡f(x)) dx\displaystyle\int \left(\dfrac{g'(x)}{g(x)}-\dfrac{f'(x)}{f(x)}\right)\big(\ln g(x)-\ln f(x)\big)\,dx equals
Aln⁡ ⁣g(x)f(x)+C\ln\!\dfrac{g(x)}{f(x)}+C
B12(ln⁡ ⁣g(x)f(x))2+C\dfrac{1}{2}\left(\ln\!\dfrac{g(x)}{f(x)}\right)^{2}+Ccorrect
Cg(x)f(x)ln⁡ ⁣g(x)f(x)+C\dfrac{g(x)}{f(x)}\ln\!\dfrac{g(x)}{f(x)}+C
Dln⁡ ⁣f(x)g(x)+C\ln\!\dfrac{f(x)}{g(x)}+C
Solution
Step 1: Substitute t=ln⁡g(x)−ln⁡f(x)=ln⁡g(x)f(x)t=\ln g(x)-\ln f(x)=\ln\dfrac{g(x)}{f(x)}. Step 2: Differentiate:
dt=(g′(x)g(x)−f′(x)f(x))dxdt=\left(\dfrac{g'(x)}{g(x)}-\dfrac{f'(x)}{f(x)}\right)dx
 which is exactly the first factor of the integrand multiplied by dxdx. Step 3: The integral collapses to
I=∫t dtI=\int t\,dt
 Step 4: Integrate and back-substitute:
I=t22+C=12(ln⁡ ⁣g(x)f(x))2+CI=\dfrac{t^{2}}{2}+C=\dfrac{1}{2}\left(\ln\!\dfrac{g(x)}{f(x)}\right)^{2}+C
 Correct answer: (2)
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