Đáp án:
\(\begin{array}{l}
25)\quad A.\, f(t) = 2t^2 - 2t\\
26)\quad D.\,f(t) = -\dfrac{2}{t^2} + \dfrac{1}{t}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
25)\quad I = \displaystyle\int\limits_0^3\dfrac{x}{1+\sqrt{1+x}}dx\\
\to I = \displaystyle\int\limits_0^3(\sqrt{1+x} - 1)dx\\
Đặt\,\,t = \sqrt{1+x}\\
\to dt = \dfrac{1}{2\sqrt{1+x}}dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 0\qquad 3\\
\overline{t\,\quad \Big|\quad 1\qquad 2}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^3(\sqrt{1+x} - 1)dx = \displaystyle\int\limits_0^32(\sqrt{1+x} -1)\sqrt{1+x}\cdot \dfrac{1}{2\sqrt{1+x}}dx\\
\to I = \displaystyle\int\limits_1^22(t-1)tdt = \displaystyle\int\limits_1^2(2t^2 - 2t)dt\\
Vậy\,\,f(t) = 2t^2 - 2t\\
26)\quad I = \displaystyle\int\limits_1^e\dfrac{\ln x}{x(\ln x + 2)^2}dx\\
Đặt\,\,t =\ln x + 2\longrightarrow t- 2 = \ln x\\
\to dt = \dfrac1xdx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 1\qquad e\\
\overline{t\,\quad \Big|\quad 2\qquad 3}\\
\text{Ta được:}\\
\quad I =\displaystyle\int\limits_1^e\dfrac{\ln x}{x(\ln x + 2)^2}dx = \displaystyle\int\limits_1^e\dfrac{(\ln x+2) - 2}{(\ln x + 2)^2}\cdot\dfrac1xdx\\
\to I = \displaystyle\int\limits_2^3\dfrac{t-2}{t^2}dt = \displaystyle\int\limits_2^3\left(\dfrac1t - \dfrac{2}{t^2}\right)dt\\
Vậy\,\,f(t) = -\dfrac{2}{t^2} + \dfrac{1}{t}
\end{array}\)
