JEE Main 2021 (Online) 25th February Evening ShiftQuestion 1
The integral $\int\frac{e^{3\log_e2x}+5e^{2\log_e2x}}{e^{4\log_ex}+5e^{3\log_ex}-7e^{2\log_ex}}dx$, $x>0$, is equal to : (where $c$ is a constant of integration)
Indefinite Integration
Solution
Concept. $e^{k\ln t}=t^k$; after simplifying, the numerator is $4\times$ the derivative of the denominator. Formulas used. (1) $e^{3\ln2x}=8x^3$, $e^{2\ln2x}=4x^2$, $e^{k\ln x}=x^k$(2) $\int\dfrac{g'}{g}dx=\ln|g|$Step 1 - Simplify. By (1): numerator $8x^3+20x^2=4x^2(2x+5)$, denominator $x^4+5x^3-7x^2=x^2(x^2+5x-7)$. So the integrand is $$\frac{4(2x+5)}{x^2+5x-7}.$$ Step 2 - Use (2). $(x^2+5x-7)'=2x+5$, so $I=4\log_e|x^2+5x-7|+c$. Answer: (B)