Bài tập 41 trang 175 SGK Toán 12 NC

Bài tập 41 trang 175 SGK Toán 12 NC

a)

\(\begin{array}{l}
\int 2 x(1 - {x^{ - 3}})dx = \int {(2x - 2{x^{ - 2}})} dx\\
 = {x^2} + 2x + C
\end{array}\)

b)

\(\begin{array}{l}
\int {\left( {8x - \frac{2}{{{x^{\frac{1}{4}}}}}} \right)dx}  = \int {\left( {8x - 2{x^{ - \frac{1}{4}}}} \right)} dx\\
 = 4{x^2} - \frac{8}{3}{x^{\frac{3}{4}}} + C
\end{array}\)

c)

Đặt \(u = {x^{\frac{3}{2}}} + 1 \Rightarrow du = \frac{3}{2}{x^{\frac{1}{2}}}dx \)

\(\Rightarrow {x^{\frac{1}{2}}}dx = \frac{2}{3}du\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\int {{x^{\frac{1}{2}}}\sin \left( {{x^{\frac{3}{2}}} + 1} \right)dx} \\
 = \frac{2}{3}\int {\sin u} du =  - \frac{2}{3}\cos u + C
\end{array}\\
{ =  - \frac{2}{3}\cos \left( {{x^{\frac{3}{2}}} + 1} \right) + C}
\end{array}\)

d) Đặt \(u = \cos (2x + 1)\)

\(\begin{array}{l}
 \Rightarrow du =  - 2\sin (2x + 1)dx\\
 \Rightarrow \sin (2x + 1)dx =  - \frac{1}{2}du
\end{array}\)

Do đó:

\(\begin{array}{l}
\int {\frac{{\sin (2x + 1)}}{{{{\cos }^2}(2x + 1)}}dx}  =  - \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} \\
 = \frac{1}{{2u}} + C = \frac{1}{{2\cos (2x + 1)}} + C
\end{array}\)

-- Mod Toán 12