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Question 103 of 104

Q.∫121x2⋅e1/x dx=\displaystyle\int_1^2 \dfrac{1}{x^2}\cdot e^{1/x}\,dx= ____.

(a) e+1\sqrt e + 1
(b) e−1\sqrt e - 1
(c) e(e−1)\sqrt e(\sqrt e - 1)
(d) e−1e\dfrac{\sqrt e - 1}{e}
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026MCQ· 2mImportance★★★★★
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Substitute u=1/xu=1/x so du=−1x2dxdu=-\dfrac{1}{x^2}dx, turning the integral into ∫eu du\int e^u\,du.

Let u=1xu=\dfrac1x, then du=−1x2dxdu=-\dfrac{1}{x^2}dx, i.e. 1x2dx=−du\dfrac{1}{x^2}dx=-du.

When x=1, u=1x=1,\ u=1; when x=2, u=12x=2,\ u=\dfrac12.

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