Question

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

Answer 1

Let $2x=t$

$\Rightarrow 2dx=dt$

$\Rightarrow I=\int e^t(\frac{2}{t}-\frac{4}{2t^2})\frac{dt}{2}$

$=\int e^t(\frac{1}{t}-\frac{1}{t^2})dt$

$=$ of form $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$

$=e^{x}f\left(x\right)+c$

$\Rightarrow I=e^t\left(\frac{1}{t}\right)+c$

$\Rightarrow I=\frac{e^{2x}}{2x}+c$