JEE PYQ: Indefinite Integration Question 33
Question 33 - 2019 (12 Jan Shift 1)
The integral $\int \left{\left(\frac{x}{e}\right)^{2x} - \left(\frac{e}{x}\right)^x\right}\log_e x,dx$ is equal to:
(1) $\frac{1}{2} - e - \frac{1}{2e^2}$
(2) $\frac{1}{2} - \frac{1}{e} - \frac{1}{2e^2}$
(3) $\frac{3}{2} - \frac{1}{e} - \frac{1}{2e^2}$
(4) $\frac{3}{2} - e - \frac{1}{2e^2}$
Show Answer
Answer: (4)
Solution
Let $\left(\frac{x}{e}\right)^x = t$, then $x(\ln x - 1) = \ln t$, and $\ln x,dx = \frac{dt}{t}$. $I = \int_1^e \left(t^2 - \frac{1}{t}\right)\frac{dt}{t}$. Evaluating the definite integral: $I = \int_1^{1/e}\left(t - \frac{1}{t^2}\right),dt = \left[\frac{t^2}{2} + \frac{1}{t}\right]_1^{1/e} = \frac{1}{2e^2} + e - \frac{1}{2} - 1 = \frac{3}{2} - e - \frac{1}{2e^2}$.
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