Question

The value of the integral $\underset{1}{\overset{2}{\int }}{e}^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx$ is _______________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{1}{\overset{2}{\int }}{e}^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx \\ \mathrm{We}\mathrm{know}, \\ \int e^x\left(f\left(x\right)+f\text{'}\left(x\right)\right)dx=e^{x}f\left(x\right)+c \\ \mathrm{Here}, \\ f\left(x\right)=\frac{1}{x} \\ \Rightarrow f\text{'}\left(x\right)=-\frac{1}{x^2} \\ \mathrm{Thus}, \\ I={\left[e^x\left(\frac{1}{x}\right)\right]}_1^2 \\ =\left[e^2\left(\frac{1}{2}\right)-e^1\left(\frac{1}{1}\right)\right] \\ =\frac{e^2}{2}-e \\ =\frac{e^2-2e}{2} \\ =\frac{e\left(e-2\right)}{2} \end{gathered}$$

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Hence, the value of the integral $\underset{1}{\overset{2}{\int }}{e}^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx$ is $\underline{\frac{e\left(e-2\right)}{2}}.$