Bài 3.15 trang 166 SBT giải tích 12

Đề bài

\(\int {x\sqrt {x - 1} dx} \) bằng

A. \({\left( {x - 1} \right)^{\dfrac{5}{2}}} + {\left( {x - 1} \right)^{\dfrac{3}{2}}} + C\)

B. \(\dfrac{2}{{15}}\left[ {3{{\left( {x - 1} \right)}^{\dfrac{5}{2}}} - 5{{\left( {x - 1} \right)}^{\dfrac{3}{2}}}} \right] + C\)

C. \(\dfrac{2}{{15}}\left[ {3{{\left( {x - 1} \right)}^{\dfrac{5}{2}}} + 5{{\left( {x - 1} \right)}^{\dfrac{3}{2}}}} \right] + C\)

D. \(\dfrac{1}{{15}}\left[ {3{{\left( {x - 1} \right)}^{\dfrac{5}{2}}} + 5{{\left( {x - 1} \right)}^{\dfrac{3}{2}}}} \right] + C\)

Lời giải chi tiết

Đặt \(t = \sqrt {x - 1}  \Rightarrow {t^2} = x - 1\) \( \Rightarrow 2tdt = dx\)

Khi đó \(\int {x\sqrt {x - 1} dx} \)\( = \int {\left( {{t^2} + 1} \right).t.2tdt} \) \( = 2\int {\left( {{t^4} + {t^2}} \right)dt} \) \( = 2\left( {\dfrac{{{t^5}}}{5} + \dfrac{{{t^3}}}{3}} \right) + C\)

\( = 2\left( {\dfrac{{{{\left( {\sqrt {x - 1} } \right)}^5}}}{5} + \dfrac{{{{\left( {\sqrt {x - 1} } \right)}^3}}}{3}} \right) + C\) \( = \dfrac{2}{{15}}\left[ {3{{\left( {x - 1} \right)}^{\dfrac{5}{2}}} + 5{{\left( {x - 1} \right)}^{\dfrac{3}{2}}}} \right] + C\).

Chọn C.