Bài tập 17 trang 161 SGK Toán 12 NC

a) Đặt \(u = \sqrt {x + 1}  \Rightarrow {u^2} = x + 1\)

\(\Rightarrow 2udu = dx.\)

Đổi cận

\(\begin{array}{l}
\int\limits_0^1 {\sqrt {x + 1} } dx = \int\limits_1^{\sqrt 2 } u .2udu\\
 = 2\int\limits_1^{\sqrt 2 } {{u^2}} du = \left. {2.\frac{{{u^3}}}{3}} \right|_1^{\sqrt 2 }\\
 = \frac{2}{3}\left( {2\sqrt 2  - 1} \right)
\end{array}\)

b) Đặt \(u = \tan x \Rightarrow du = \frac{{dx}}{{{{\cos }^2}x}}\)

\(\int \limits_0^{\frac{\pi }{4}} \frac{{\tan x}}{{{{\cos }^2}x}}dx = \int\limits_0^1 {udu}  = \left. {\frac{{{u^2}}}{2}} \right|_0^1 = \frac{1}{2}\)

c) Đặt \(u = 1 + {t^4} \Rightarrow du = 4{t^3}dt \)

\(\Rightarrow {t^3}dt = \frac{{du}}{4}\)

\(\begin{array}{l}
\int\limits_0^1 {{t^3}} {\left( {1 + {t^4}} \right)^3}dt = \frac{1}{4}\int {{u^3}} du\\
 = \left. {\frac{1}{4}.\frac{{{u^4}}}{4}} \right|_1^2 = \frac{1}{{16}}(16 - 1) = \frac{{15}}{{16}}
\end{array}\)

d) Đặt \(u = {x^2} + 4 \Rightarrow du = 2xdx\)

\(\Rightarrow xdx = \frac{1}{2}du\)

\(\begin{array}{l}
\int\limits_0^1 {\frac{{5x}}{{{{\left( {{x^2} + 4} \right)}^2}}}} dx = \frac{5}{2}\int\limits_4^5 {\frac{{du}}{{{u^2}}}} \\
 = \left. {\frac{5}{2}\left( { - \frac{1}{u}} \right)} \right|_4^5 = \frac{1}{8}
\end{array}\)

e) Đặt \(u = \sqrt {{x^2} + 1}  \Rightarrow {u^2} = {x^2} + 1 \)

\(\Rightarrow udu = xdx\)

\(\begin{array}{l}
\int\limits_0^{\sqrt 3 } {\frac{{4x}}{{\sqrt {{x^2} + 1} }}} dx = 4\int\limits_1^2 {\frac{{udu}}{u}} \\
 = \left. {4u} \right|_1^2 = 4
\end{array}\)

f) Đặt \(u = 1 - \cos 3x \Rightarrow du = 3\sin 3xdx \)

\(\Rightarrow \sin 3xdx = \frac{1}{3}du\)

\(\begin{array}{l}
\int\limits_0^{\frac{\pi }{6}} {\left( {1 - \cos 3x} \right)} \sin 3xdx\\
 = \frac{1}{3}\int\limits_0^1 {udu}  = \left. {\frac{{{u^2}}}{6}} \right|_0^1 = \frac{1}{6}
\end{array}\)

-- Mod Toán 12 HỌC247