Giải bài 5 trang 113 - SGK môn Giải tích lớp 12

Gợi ý:

a) Dùng phương pháp đổi biến: Đặt $1+3x =t$

b) Tách $\dfrac{x^3-1}{x^2-1}=x+\dfrac 1 {x+1}$

c) Dùng phương pháp tích phân từng phần

a) $\int\limits_{0}^{1}{{{\left( 1+3x \right)}^{\frac{3}{2}}}dx}$

Đặt $1+3x=t\Rightarrow dt=3dx\Rightarrow dx=\dfrac{dt}{3}$

Đổi cận

$\begin{aligned} & \int\limits_{0}^{1}{{{\left( 1+3x \right)}^ {\frac{3}{2}}}dx}=\dfrac{1}{3}\int\limits_{1}^{4}{{{t}^{\frac{3}{2}}}dt} \\ & =\dfrac{2}{15}{{t}^{\frac{5}{2}}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 4 \\ \end{smallmatrix}} \right.=\dfrac{64-2}{15}=\dfrac{62}{15} \\ \end{aligned}$

b) 

$\begin{aligned} & \int\limits_{0}^{\frac{1}{2}}{\dfrac{{{x}^{3}}-1} {{{x}^{2}}-1}dx}=\int\limits_{0}^{\frac{1}{2}}{\left( x+\dfrac{1}{x+1} \right)dx} \\ & =\left( \dfrac{{{x}^{2}}}{2}+\ln \left| x+1 \right| \right)\left| _{\begin{smallmatrix} \\ \\\\ \\ \\ 0 \end{smallmatrix}}^{\frac{1}{2}} \right. \\ & =\dfrac{1}{8}+\ln \dfrac{3}{2} \\ \end{aligned}$

c) $\int\limits_{1}^{2}{\dfrac{\ln \left( 1+x \right)}{{{x}^{2}}}dx}$

Đặt $\left\{ \begin{aligned} & \ln \left( 1+x \right)=u \\ & \dfrac{dx}{{{x}^{2}}}=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\dfrac{dx}{1+x} \\ & v=-\dfrac{1}{x} \\ \end{aligned} \right.$

$\begin{aligned} & \int\limits_{1}^{2}{\dfrac{\ln \left( 1+x \right)}{{{x}^{2}}}dx}=-\dfrac{\ln \left( 1+x \right)}{x}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right.+\int\limits_{1}^{2}{\dfrac{1}{x\left( x+1 \right)}dx} \\ & =-\dfrac{1}{2}\ln 3+ln2+\int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{\left( x+1 \right)} \right)dx} \\ & =-\dfrac{1}{2}\ln 3+ln2+\ln \left| \dfrac{x}{x+1} \right|\,\,\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right. \\ & =-\dfrac{1}{2}\ln 3+ln2+\ln \dfrac{2}{3}-\ln \dfrac{1}{2} \\ & =3\ln 2+\dfrac{3}{2}\ln 3 \\ \end{aligned}$