Question
Evaluate the definite integral: $\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
✓
Answer
We have,
$ I =\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
$I =\int_1^2 \frac{1}{x} \cdot e^{2 x}-\int_1^2 \frac{1}{2 x^2} \cdot e^{2 x} d x$
$\Rightarrow I = I _1- I _2$
Now, $I _1=\int_1^2 \frac{1}{x} e^{2 x} \ ($By parts we have$)$
$\Rightarrow I _1=\left[\frac{1}{x}\right]_1^2 \cdot \int_1^2 e^{2 x} d x-\int_1^2-\frac{1}{x^2} \frac{e^{2 x}}{2} d x$
$\Rightarrow I _1=\left[\frac{1}{x} \cdot \frac{e^{2 x}}{2}\right]_1^2+\int_1^2 \frac{1}{2 x^2} e^{2 x} d x$
$\Rightarrow I _1=\left[\frac{1}{2 x} e^{2 x}\right]_1^2+ I _2$
As, $I = I _1- I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2- I _2+ I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2=\frac{1}{2}\left[\frac{1}{2} e^4-e^2\right]$
$\Rightarrow I =\frac{1}{4} e ^2\left( e ^2-1\right)$
$ I =\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
$I =\int_1^2 \frac{1}{x} \cdot e^{2 x}-\int_1^2 \frac{1}{2 x^2} \cdot e^{2 x} d x$
$\Rightarrow I = I _1- I _2$
Now, $I _1=\int_1^2 \frac{1}{x} e^{2 x} \ ($By parts we have$)$
$\Rightarrow I _1=\left[\frac{1}{x}\right]_1^2 \cdot \int_1^2 e^{2 x} d x-\int_1^2-\frac{1}{x^2} \frac{e^{2 x}}{2} d x$
$\Rightarrow I _1=\left[\frac{1}{x} \cdot \frac{e^{2 x}}{2}\right]_1^2+\int_1^2 \frac{1}{2 x^2} e^{2 x} d x$
$\Rightarrow I _1=\left[\frac{1}{2 x} e^{2 x}\right]_1^2+ I _2$
As, $I = I _1- I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2- I _2+ I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2=\frac{1}{2}\left[\frac{1}{2} e^4-e^2\right]$
$\Rightarrow I =\frac{1}{4} e ^2\left( e ^2-1\right)$
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