Evaluate the integral in Exercise: ^ 2 _ 1 (1 x -1 2x^ 2 )e^ 2x d… - Vidyadip

Question

Evaluate the integral in Exercise:
$\int^{2}_{1}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$

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Answer

$\text{Let}$
$\text{I}=\int_{1}^{2}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$
$=\int^{2}_{1}\frac{1}{\text{x}}\text{e}^{2\text{x}}\text{dx}+\int\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\bigg[\frac{1}{\text{x}}\frac{\text{e}^{2\text{x}}}{2}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{\text{x}^{2}}\frac{\text{e}^{2\text{x}}}{2}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$ [integrating by parts]
$=\frac{1}{2}\bigg[\frac{\text{e}^{2\text{x}}}{\text{x}}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{2\text{x}^{2}}{\text{e}^{2\text{x}}}{}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\frac{1}{2}\bigg[\frac{\text{e}^{4}}{2}-\frac{\text{e}^{2}}{1}\bigg]=\frac{1}{4}\big(\text{e}^{4}-2\text{e}^{2}\big)=\frac{1}{4}\big(\text{e}^{2}-2\big)$

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