Đáp án: $B$
Giải thích các bước giải:
Đặt $\sqrt{x-1}=u$
$\to x-1=u^2$
$\to d(x-1)=du^2$
$\to dx=2udu$
Mặt khác $x-1=u^2\to x=u^2+1$
$\to F(x)=\displaystyle\int(u^2+1)^2u\cdot 2udu$
$\to F(x)=2\displaystyle\int(u^2+1)^2u^2du$
$\to F(x)=2\displaystyle\int(u^6+2u^4+u^2)du$
$\to F(x)=2(\dfrac{u^7}{7}+\dfrac{2u^5}{5}+\dfrac{u^3}{3})+C$
$\to F(x)=\dfrac27u^7+\dfrac25u^5+\dfrac13u^3+C$
Lại có:
$F(x)=ax^2(x-1)\sqrt{x-1}+bx(x-1)^2\sqrt{x-1}+c(x-1)^3\sqrt{x-1}+C$
$\to F(x)=a(u^2+1)^2u^3+b(u^2+1)u^5+cu^7+C$
$\to F(x)=a(u^4+2u^2+1)u^3+b(u^7+u^5)+cu^7+C$
$\to F(x)=a(u^7+2u^5+u^3)+b(u^7+u^5)+cu^7+C$
$\to F(x)=(a+b+c)u^7+(2a+b)u^5+au^3+C$
$\to\begin{cases}a+b+c=\dfrac27\\ 2a+b=\dfrac25\\ c=\dfrac13\end{cases}$
$\to B$
