Question

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

Answer 1

Step 1: Substitute the appropriate values in the integral.

Given: ${\int }_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}dx$

Let $2x=t\Rightarrow x=\frac{t}{2}$

Differentiate the above equation we get $2dx=dt\Rightarrow dx=\frac{dt}{2}$

When $x=1\Rightarrow t=2$ and $x=2\Rightarrow t=4$

Now put the value of $2x,x$ and $dx$in the given expression.

$={\int }_2^4\left(\frac{1}{{\displaystyle \frac{t}{2}}}-\frac{1}{2{\displaystyle \frac{t^2}{4}}}\right)e^t\frac{dt}{2}$

Step 2: Simplify the above expression

$$\begin{gathered} ={\int }_2^4\left(\frac{2}{t}-\frac{2}{t^2}\right)e^t\frac{dt}{2} \\ ={\int }_2^4\left(\frac{1}{t}-\frac{1}{t^2}\right)e^{t}dt...\left(1\right) \end{gathered}$$

Now let $f\left(t\right)=\frac{1}{t}$and $f\text{'}\left(t\right)=\frac{-1}{t^2}$

Now put values of $f\left(t\right),f\text{'}\left(t\right)$ in $\left(1\right)$

$={\int }_2^4\left(f\left(t\right)+f\text{'}\left(t\right)\right)e^{t}dt$

Step 3: Solve the integral

$$\begin{gathered} ={[e^t·f(t\left)\right]}_2^4\mathbf{}\left[\because \int e^t\left[f\right(t)+f\text{'}(t\left)\right]\mathrm{dt}=e^{t}f\left(t\right)\right] \\ ={[e^t·\frac{1}{t}]}_2^4 \\ =\left[e^4·\frac{1}{4}-e^2·\frac{1}{2}\right] \\ =e^2·\frac{\left(e^2-2\right)}{4} \end{gathered}$$

Hence the required answer is $\frac{e^2\left(e^2-2\right)}{4}$