Question

Evaluate the integral ${\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$ using substitution.

Answer 1

${\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$
Let $2x=t\Rightarrow 2dx=dt$
When $x=1,t=2$ and when $x=2,t=4$
$\therefore {\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx={\int }_2^4\frac{1}{2}(\frac{2}{t}-\frac{2}{t^2})e^{t}dt$
$={\int }_2^4(\frac{1}{t}-\frac{1}{t^2})e^{t}dt$
Let $\frac{1}{t}=f\left(t\right)$
Then, $f^{\prime }\left(t\right)=-\frac{1}{t^2}$
$\Rightarrow {\int }_2^4(\frac{1}{t}-\frac{1}{t^2})e^{t}dt={\int }_2^4{e}^t\left[f\right(t)+f^{\prime }(t\left)\right]dt$
$=\left[e^{t}f\right(t){]}_2^4$
$={[e^t\cdot \frac{1}{t}]}_2^4$
$={\left[\frac{e^t}{t}\right]}_2^4$
$=\frac{e^4}{4}-\frac{e^2}{2}$
$=\frac{e^2(e^2-2)}{4}$