Q112 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

$\int e^x\left(\log x+\frac{1}{x^2}\right) d x=$

  • A
    $e^x\left(\log x+\frac{1}{x}\right)+C$
  • Correct answer
    B
    $\mathrm{e}^{\mathrm{x}}\left(\log \mathrm{x}-\frac{1}{\mathrm{x}}\right)+\mathrm{C}$
  • C
    $e^x\left(\log x+\frac{2}{x}\right)+C$
  • D
    $\mathrm{e}^{\mathrm{x}}\left(\log \mathrm{x}-\frac{2}{\mathrm{x}}\right)+\mathrm{C}$

Solution

$\int e^x\left(\log x+\frac{1}{x^2}\right) d x$ $\begin{aligned} & =\int e^x\left(\log x-\frac{1}{x}+\frac{1}{x}+\frac{1}{x^2}\right) d x \\ & =e^x\left(\log x-\frac{1}{x}\right)+C\end{aligned}$

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

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