$\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
✓
Answer
Let $I=\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$ Put $2 x=t \quad \therefore 2 d x=d t$ $\therefore d x=\frac{d t}{2}$ and $x=\frac{t}{2}$ When $x=1, t=2$ When $x=2, t=4$ $ \therefore I=\int_2^4 e^t\left(\frac{2}{t}-\frac{2}{t^2}\right) \frac{d t}{2}=\frac{1}{2} \int_2^4 e^t\left(\frac{2}{t}-\frac{2}{t^2}\right) d t $ Let $f(t)=\frac{2}{t}$ Then $f^{\prime}(t)=2\left(-\frac{1}{t^2}\right)=\frac{-2}{t^2}$ $ \begin{aligned} \therefore I & =\frac{1}{2} \int_2^4 e^t\left[f(t)+f^{\prime}(t)\right] d t \\ & =\frac{1}{2}\left[e^t \cdot f(t)\right]_2^4=\frac{1}{2}\left[e^t \cdot \frac{2}{t}\right]_2^4 \\ & =\frac{1}{2}\left[e^4 \times \frac{2}{4}-e^2 \times \frac{2}{2}\right] \\ & =\frac{e^4}{4}-\frac{e^2}{2} . \end{aligned} $
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