Đáp án:
\(\begin{array}{l}27)\quad B.\, I = \displaystyle\int\limits_0^1(1-u)e^{-u}du\\
28)\quad A.\, I = \dfrac23\displaystyle\int\limits_1^2tdt
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
27)\quad I = \displaystyle\int\limits_1^e\dfrac{1 - \ln x}{x^2}dx\\
Đặt\,\,u = \ln x\longrightarrow x = e^u\\
\to du = \dfrac1xdx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 1\qquad e\\
\overline{u\quad \Big|\quad 0\qquad 1}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_1^e\dfrac{1 - \ln x}{x^2}dx = \displaystyle\int\limits_1^e\dfrac{1 - \ln x}{x}\cdot \dfrac1xdx\\
\to I = \displaystyle\int\limits_0^1\dfrac{1 - u}{e^u}du = \displaystyle\int\limits_0^1(1-u)e^{-u}du\\
28)\quad I = \displaystyle\int\limits_1^e\dfrac{\sqrt{1 + 3\ln x}}{x}dx\\
Đặt\,\,t = \sqrt{1 + 3\ln x}\longrightarrow t^2 = 1 + 3\ln x\\
\to du = \dfrac{3}{2}\cdot \dfrac{1}{x\sqrt{1 + 3\ln x}}dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 1\qquad e\\
\overline{u\quad \Big|\quad 1\qquad 2}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_1^e\dfrac{\sqrt{1 + 3\ln x}}{x}dx = \dfrac23\displaystyle\int\limits_1^e(1+3\ln x)\cdot \dfrac32\cdot \dfrac{1}{x\sqrt{1+3\ln x}}dx\\
\to I = \dfrac23\displaystyle\int\limits_1^2t^2dt\\
\to I = \dfrac29t^3\Bigg|_1^2\\
\to I = \dfrac{14}{9}
\end{array}\)
