Bài 5 (Trang 113 SGK Toán Giải tích 12):

Tính các tích phân sau:

a) ${\int }_0^1{(1+3\mathrm{x})}^{\frac{3}{2}}\mathrm{dx}$ ;

b)${\int }_0^{\frac{1}{2}}\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^2-1}\mathrm{dx}$ ;

c) ${\int }_1^2\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}$ .

Hướng dẫn giải:

a)${\int }_0^1{(1+3\mathrm{x})}^{\frac{3}{2}}\mathrm{dx}=\frac{1}{2}.\frac{2}{5}{(1+3\mathrm{x})}^{\frac{5}{2}}{|}_0^1=\frac{2}{15}(4^{\frac{5}{2}}-1).$

$\mathrm{b}) {\int }_0^{\frac{1}{2}}\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^2-1}\mathrm{dx} = {\int }_0^{\frac{1}{2}}\frac{{\mathrm{x}}^2+\mathrm{x}+1}{\mathrm{x}+1}\mathrm{dx} = {\int }_0^{\frac{1}{2}}(\mathrm{x}+\frac{1}{\mathrm{x}+1})\mathrm{dx} = (\frac{{\mathrm{x}}^2}{2}+\ln \left|\mathrm{x}+1\right|){|}_0^{\frac{1}{2}} =\frac{1}{8}+\ln \frac{3}{2}$

c) Đặt $\left\{\begin{array}{l}\mathrm{u}=\ln (1+\mathrm{x})\\ \mathrm{dv}=\frac{1}{{\mathrm{x}}^2}\mathrm{dx}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{du}=\frac{\mathrm{dx}}{1+\mathrm{x}}\\ \mathrm{v}=-\frac{1}{\mathrm{x}}\end{array}\right.\right.$

${\int }_1^2\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx} = -\frac{\ln (1+\mathrm{x})}{\mathrm{x}}{|}_1^2 +{\int }_1^2\frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)} = -\frac{\ln (1+\mathrm{x})}{\mathrm{x}}{|}_1^2 + \ln \frac{\mathrm{x}}{\mathrm{x}+1}{|}_1^2 = 3\ln \frac{2\sqrt{3}}{3}$