Question:

$\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$

Solution:

$\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$

Let $2 x=t \Rightarrow 2 d x=d t$

When $x=1, t=2$ and when $x=2, t=4$

$\therefore \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x=\frac{1}{2} \int_{2}^{4}\left(\frac{2}{t}-\frac{2}{t^{2}}\right) e^{t} d t$

$=\int_{2}^{+}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) e^{t} d t$

Let $\frac{1}{t}=f(t)$

Then, $f^{\prime}(t)=-\frac{1}{t^{2}}$

$\Rightarrow \int_{2}^{1}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) e^{t} d t=\int_{2}^{1} e^{t}\left[f(t)+f^{\prime}(t)\right] d t$

$=\left[e^{t} f(t)\right]_{2}^{4}$

$=\left[e^{t} \cdot \frac{2}{t}\right]_{2}^{4}$

$=\left[\frac{e^{t}}{t}\right]_{2}^{4}$

$=\frac{e^{t}}{4}-\frac{e^{2}}{2}$

$=\frac{e^{2}\left(e^{2}-2\right)}{4}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.