Question

Evaluate$\int e^x\left(\frac{2}{x}-\frac{2}{x^2}\right)dx=$

• A. $\left(\frac{e^x}{x}\right)+c$
• B. $\left(\frac{e^x}{2x^2}\right)+c$
• C. $\left(\frac{2e^x}{x}\right)+c$
• D. $\left(\frac{2e^x}{x^2}\right)+c$

Answer 1

The correct option is C

$\left(\frac{2e^x}{x}\right)+c$

Explanation for the correct option:

Finding the integral:

$$\begin{gathered} \int e^x\left(\frac{2}{x}-\frac{2}{x^2}\right)dx=2\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx......\text{Taking 2 common} \\ =2\int e^x\frac{1}{x}dx-2\int \frac{e^x}{x^2}dx \\ \text{By using integration by parts} \\ =2·\left[\frac{1}{x}·\int e^x-\int \frac{d\frac{1}{x}}{dx}\int e^{x}dx\right]-2\int \frac{e^x}{x^2}dx+c \\ =2·\left[\frac{1}{x}·e^x-\int -\frac{1}{x^2}{e}^{x}dx\right]-2\int \frac{e^x}{x^2}dx+c \\ =2·\frac{1}{x}·e^x+2\int \frac{e^x}{x^2}dx-2\int \frac{e^x}{x^2}dx+c \\ =\frac{2e^x}{x}+c \end{gathered}$$ ${\mathbf{\int }}_{}^{}\mathbf{u}\mathbf{.}\mathbf{vdx}\mathbf{=}\mathbf{u}{\mathbf{\int }}_{}^{}\mathbf{vdx}\mathbf{-}{\mathbf{\int }}_{}^{}\left(u\text{'}{\int }_{}^{}\mathrm{vdx}\right)$

Hence, option (C) is correct.