Giải bài 5 trang 145 SGK giải tích nâng cao 12

a) $\int{\dfrac{9{{x}^{2}}}{\sqrt{1-{{x}^{3}}}}dx}$

Đặt $u=1-{{x}^{3}}$ $\Rightarrow du=-3{{x}^{2}}dx\Leftrightarrow 3{{x}^{2}}dx=-du$

$\int{\dfrac{9{{x}^{2}}}{\sqrt{1-{{x}^{3}}}}dx}=-\int{\dfrac{3du}{\sqrt{u}}} \\ =-6\sqrt{u}+C=-6\sqrt{1-{{x}^{3}}}+C$

b) $\int{\dfrac{1}{\sqrt{5x+4}}dx}$

Đặt $u=5x+4$ $\Rightarrow du=5dx\Leftrightarrow dx=\dfrac{du}{5}$

$\int{\dfrac{1}{\sqrt{5x+4}}dx}=\dfrac{1}{5}\int{\dfrac{du}{\sqrt{u}}=\dfrac{2}{5}\sqrt{u}+C} \\ =\dfrac{2}{5}\sqrt{5x+4}+C$

c) $\int{x\sqrt[4]{1-{{x}^{2}}}dx}$

Đặt $u=1-{{x}^{2}}$ $\Rightarrow du=-2xdx\Leftrightarrow xdx=\dfrac{-du}{2}$

$\int{x\sqrt[4]{1-{{x}^{2}}}dx}=-\dfrac{1}{2}\int{\sqrt{u}du}=-\dfrac{1}{3}\sqrt{{{u}^{3}}}+C \\ =-\dfrac{1}{3}\sqrt{{{\left( 1-{{x}^{2}} \right)}^{3}}}+C$

d) $\int{\dfrac{1}{\sqrt{x}{{\left( 1+\sqrt{x} \right)}^{2}}}dx}$

Đặt $u=1+\sqrt{x}$  $\Rightarrow du=\dfrac{dx}{2\sqrt{x}}\Leftrightarrow 2du=\dfrac{dx}{\sqrt{x}}$

$\int{\dfrac{1}{\sqrt{x}{{\left( 1+\sqrt{x} \right)}^{2}}}dx}=\int{\dfrac{2du}{{{u}^{2}}}=-\dfrac{2}{u}+C} \\ =-\dfrac{2}{1+\sqrt{x}}+C$