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What is the value of \(\displaystyle\int_1^2 e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \ ?\)1. \(e\left(\dfrac{e}{2}-1\right)\)2. e(e - 1)3. \(e-\dfrac{1}{e}\)4. 0 |
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Answer» Correct Answer - Option 1 : \(e\left(\dfrac{e}{2}-1\right)\) Concept: Integration by parts: ∫u v dx = u∫v dx −∫u' (∫v dx) dx ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.
Calculation: Let, \(\rm I=\displaystyle\int_1^2 e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \) \(\rm =\displaystyle\int_1^2 (\dfrac{e^x}{x})dx -\displaystyle\int_1^2 \left( \dfrac{1}{x^2}\right)dx \) \(\rm =[\frac1 x\displaystyle\int_1^2 {e^x}dx]_1^2+\displaystyle\int_1^2 \left( \dfrac{e^x}{x^2}\right)dx -\displaystyle\int_1^2 \left( \dfrac{e^x}{x^2}\right)dx \) \(\rm =[\frac1 x( e^x)]_1^2\) \(\rm =\frac1 2( e^2)-e\) \(\rm =e\left(\dfrac{e}{2}-1\right)\) Hence, option (3) is correct. |
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