Q.1 (vii) displaystyle∫₁² 1/x²· e^(1/x) dx = _____. A. √(e) + 1 B. √(e) - 1 C. √(e)(√(e) - 1) D. √(e) - 1/e

Answer: C. √(e)(√(e) - 1) Answer: (c) √(e)(√(e) - 1) Put 1/x = t , -1/x²dx = dt . When x = 1,\ t = 1 ; when x = 2,\ t = 1/2 . Integral = -displaystyle∫₁^(1/2) e^t dt = ∫_(1/2)¹ e^t dt = e - e^(1/2) = √(e)(√(e) - 1) .

\(\displaystyle\int_1^2 \frac{1}{x^2}\cdot e^{\frac{1}{x}}\,dx =\) _____.

Q.1 (vii) \(\displaystyle\int_1^2 \frac{1}{x^2}\cdot e^{\frac{1}{x}}\,dx =\) _____.

A. \(\sqrt{e} + 1\)

B. \(\sqrt{e} - 1\)

C. \(\sqrt{e}(\sqrt{e} - 1)\)

D. \(\dfrac{\sqrt{e} - 1}{e}\)

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: C. \(\sqrt{e}(\sqrt{e} - 1)\)

Answer: (c) \(\sqrt{e}(\sqrt{e} - 1)\)

Put \(\dfrac{1}{x} = t\), \(-\dfrac{1}{x^2}dx = dt\). When \(x = 1,\ t = 1\); when \(x = 2,\ t = \dfrac{1}{2}\). Integral \(= -\displaystyle\int_1^{1/2} e^t\,dt = \int_{1/2}^{1} e^t\,dt = e - e^{1/2} = \sqrt{e}(\sqrt{e} - 1)\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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\(\displaystyle\int_1^2 \frac{1}{x^2}\cdot e^{\frac{1}{x}}\,dx =\) _____.

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