Value of _1^4 (_e x)/x^2 dx is Option 1: 3 - _e 4 Option 2: 1/4(3 - _e 4) Option 3: 1/4_e 4 Option 4: 1/4_e 1/4

Answer: 1/4(3 - _e 4). Solution: Integrate by parts with u=_e x and dv=x^(-2)dx, so v=-1x. This gives [-(_e x)/x]_1^4+_1^4dx/x^2=-(_e 4)/4+34=14(3-_e 4).

Calculus

Integrals

Medium

core

Value of $\int_{1}^{4} \frac{\log_e x}{x^2}\, dx$ is

$3 - \log_e 4$

$\frac{1}{4}(3 - \log_e 4)$

$\frac{1}{4}\log_e 4$

$\frac{1}{4}\log_e \frac{1}{4}$

✅ Correct Option: 2

Integrate by parts with $u=\log_e x$ and $dv=x^{-2}dx$ , so $v=-\frac1x$ . This gives $\left[-\frac{\log_e x}{x}\right]_1^4+\int_1^4\frac{dx}{x^2}=-\frac{\log_e 4}{4}+\frac34=\frac14(3-\log_e 4)$ .

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Match List I with List II

LIST I	LIST II
A. $\int \frac{\sin x}{1 + \cos x} \, dx$	I. $e^{\tan^{-1} x} + C$
B. $\int \frac{1}{1 - \tan x} \, dx$	II. $\log(\log x + 1) + C$
C. $\int \frac{e^{\tan^{-1} x}}{1 + x^2} \, dx$	III. $-\log\lvert 1+\cos x \rvert + C$
D. $\int \frac{1}{x + x \log x} \, dx$	IV. $\frac{x}{2} - \frac{1}{2}\log\lvert \cos x - \sin x \rvert + C$

Choose the correct answer from the options given below:

Value of ∫14log⁡exx2 dx\int_{1}^{4} \frac{\log_e x}{x^2}\, dx∫14​x2loge​x​dx is

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Value of $\int_{1}^{4} \frac{\log_e x}{x^2}\, dx$ is