Q113 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

If $\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x=f(x)+c$, then $f(1)-f(-1)=$

  • A
    $e^2-\frac{1}{e^2}$
  • Correct answer
    B
    $e^2+\frac{1}{e^2}$
  • C
    $e+\frac{1}{e}$
  • D
    $e-\frac{1}{e}$

Solution

$\begin{aligned} & \mathrm{I}=\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x \\ & =\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^2}\right) e^{\left(x+\frac{1}{x}\right)} d x \\ & \text { Using } \int\left[f(x)+x f^{\prime}(x)\right] d x=x f(x)+c \\ & \mathbb{I}=x e^{\left(x+\frac{1}{x}\right)}+c \\ & f(1)-f(-1)=e^2+\frac{1}{e^2}\end{aligned}$

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

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