Q187 / 270 · NDA 2010 (Phase 2) · Mathematics

What is the value of $\int_1^2 e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$

  • Correct answer
    A
    $e\left(\frac{e}{2}-1\right)$
  • B
    $e(e-1)$
  • C
    $e-\frac{1}{e}$
  • D
    0

Solution

$\begin{aligned} & \int_1^2\left(\frac{e^x}{x}-\frac{e^x}{x^2}\right) d x \\ & =\int_1^2 \frac{e^x}{x} d x-\int_1^2 \frac{e^x}{x^2} d x \\ & =\left[\frac{1}{x} \cdot e^x\right]_1^2-\int_1^2-\frac{1}{x^2} \cdot e^x d x-\int_1^2 \frac{e^x}{x^2} d x \\ & =\frac{1}{2} e^2-e^1=e\left(\frac{e}{2}-1\right)\end{aligned}$

Hence, the correct answer is option: 'A'.

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