a) Đặt \(x - 9 = t\) \( \Rightarrow dx = dt\)
Khi đó \(\int {{{\left( {x - 9} \right)}^4}dx} \) \( = \int {{t^4}dt} = \dfrac{{{t^5}}}{5} + C\)\( = \dfrac{{{{\left( {x - 9} \right)}^5}}}{5} + C\)
Vậy \(F\left( x \right) = \dfrac{{{{\left( {x - 9} \right)}^5}}}{5} + C\)
b) Đặt \(2 - x = t \Rightarrow dx = - dt\)
Khi đó \(\int {\dfrac{1}{{{{\left( {2 - x} \right)}^2}}}dx} = \int {\dfrac{{ - dt}}{{{t^2}}}} \) \( = \dfrac{1}{t} + C = \dfrac{1}{{2 - x}} + C\)
Vậy \(F(x) = \dfrac{1}{{2 - x}} + C\)
c) Đặt \(\sqrt {1 - {x^2}} = t \Rightarrow 1 - {x^2} = {t^2}\) \( \Rightarrow - 2xdx = 2tdt \Leftrightarrow xdx = - tdt\)
Khi đó \(\int {\dfrac{x}{{\sqrt {1 - {x^2}} }}dx} = \int {\dfrac{{ - tdt}}{t}} = \int { - dt} \) \( = - t + C = - \sqrt {1 - {x^2}} + C\)
Vậy \(F(x) = - \sqrt {1 - {x^2}} + C\)
d) Đặt \(\sqrt {2x + 1} = t \Rightarrow 2x + 1 = {t^2}\) \( \Rightarrow 2dx = 2tdt \Rightarrow dx = tdt\)
Khi đó \(\int {\dfrac{1}{{\sqrt {2x + 1} }}dx} = \int {\dfrac{{tdt}}{t}} = \int {dt} \) \( = t + C = \sqrt {2x + 1} + C\)
Vậy \(F(x) = \sqrt {2x + 1} + C\)
