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JEE Previous Year Question

$\int_{1}^{4} (x + x + \frac{1}{x}) dx - \int_{1}^{2 log 2} dx =$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $\frac{79}{6}$
B $\frac{643}{6}$
C $\frac{321}{5}$
D $64$

Correct answer: A

Solution

(a) \int_{1}^{4} (x + x + \frac{1}{x}) dx - \int_{1}^{2 log 2} dx \Rightarrow (\frac{x ^{2}}{2} + \frac{x ^{3/2}}{3/2} + ln | x |)_{1}^{4} - [x]_{1}^{2 log 2} \Rightarrow \frac{15}{2} + \frac{14}{3} + 1 = \frac{79}{6}

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