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.

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