Bài tập 3.3 trang 164 SBT Toán 12

Bài tập 3.3 trang 164 SBT Toán 12

a) \(f\left( x \right) = {\left( {x - 9} \right)^4}\)${\mathrm{}}^{}$

\(\mathop \smallint \nolimits {(x - 9)^4}dx = \frac{1}{5}{(x - 9)^5} + C\)

b)  \(f\left( x \right) = \frac{1}{{{{\left( {2 - x} \right)}^2}}}\)

Đặt u = 2−x. Ta có du = −dx

\(\mathop \smallint \nolimits \frac{1}{{{{\left( {2 - x} \right)}^2}}}dx =  - \mathop \smallint \nolimits \frac{{du}}{{{u^2}}} = \frac{1}{u} + C = \frac{1}{{2 - x}} + C\)

c) \(f\left( x \right) = \frac{x}{{\sqrt {1 - {x^2}} }}\)

Đặt \(\sqrt {1 - {x^2}}  = u \Leftrightarrow 1 - {x^2} = {u^2}\)

\( \Rightarrow  - 2xdx = 2udu \Leftrightarrow xdx =  - udu\)

\(\mathop \smallint \nolimits \frac{x}{{\sqrt {1 - {x^2}} }}dx =  - \mathop \smallint \nolimits \frac{{udu}}{u} =  - \mathop \smallint \nolimits^ du =  - u + C =  - \sqrt {1 - {x^2}}  + C\)

d) \(f\left( x \right) = \frac{1}{{\sqrt {2x + 1} }}\)

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

\( \Rightarrow 2dx = 2udu \Leftrightarrow dx = udu\)

\(\mathop \smallint \nolimits \frac{1}{{\sqrt {2x + 1} }}dx = \mathop \smallint \nolimits \frac{{udu}}{u} = \mathop \smallint \nolimits du = u + C = \sqrt {2x + 1}  + C\)

-- Mod Toán 12