Question

The value of $\int e^{2x}(\frac{1}{x}-\frac{1}{2x^2})dx$ is

• A. $\frac{e^{2x}}{2}+c$
• B. $\frac{e^{2x}}{2x}+c$
• C. $\frac{e^{2x}}{3x}+c$
• D. $\frac{e^{2x}}{x}+c$

Answer 1

The correct option is B $\frac{e^{2x}}{2x}+c$

$I=\int e^{2x}(\frac{1}{x}-\frac{1}{2x^2})dx$

Let $t=2x,\Rightarrow dt=2dx$

$\therefore I=\frac{e^t}{2}(\frac{2}{t}-\frac{2}{t^2})dt$

This can be compared with the standard form that $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^x\times f\left(x\right)$

Thus, $I=\frac{e^t}{2}\times \frac{2}{t}$

$I=\frac{e^t}{t}+c$

Replacing $t$ by $2x,$ we get $I$ to be $\frac{e^{2x}}{2x}+c$