Giải bài 2 trang 100 – SGK môn Giải tích lớp 12

$\begin{aligned} a)\,\int{\dfrac{x+\sqrt{x}+1}{\sqrt[3]{x}}\,}dx&=\int{\left( {{x} ^{\frac{2}{3}}}+{{x}^{\frac{1}{6}}}+1 \right)dx} \\ & =\dfrac{3}{5}{{x}^{\frac{5}{3}}}+\dfrac{6}{7}{{x}^{\frac{7}{6}}}+x+C \\ \end{aligned}$

$\begin{aligned} b)\, \int{\dfrac{{{2}^{x}}-1}{{{e}^{x}}}}\,dx&=\int{\left[ {{\left( \dfrac{2}{e} \right)}^{x}}-{{e}^{-x}} \right]dx} \\ & =\dfrac{1}{\ln \dfrac{2}{e}}{{\left( \dfrac{2}{e} \right)} ^{x}}+{{e}^{-x}}+C \\ & =\dfrac{{{2}^{x}}}{\left( \ln 2-1 \right){{e}^{x}}}+\dfrac{1}{{{e}^{x}}}+C\\ \end{aligned}$

$\begin{aligned} c)\,\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}\,dx}&=\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx\, \\ & =\int{\left( \dfrac{1}{{{\cos }^{2}}x}+\dfrac{1}{{{\sin }^{2}}x} \right)\,dx} \\ & =\tan x-\cot x+C\\ \end{aligned}$

$\begin{aligned} d)\,\int{\sin 5x.\cos 3x\,dx}&=\int{\dfrac{1}{2}\left( \sin 8x+\sin 2x \right)dx} \\ & =-\dfrac{1}{16}\cos 8x-\dfrac{1}{4}\cos 2x+C \\ \\ \end{aligned}$

$\begin{aligned} e)\,\int{{{\tan }^{2}}x\,dx}&=\int{\left( \dfrac{1}{{{\cos }^{2}}x}-1 \right)dx} \\ & =\tan x-x+C\\ \end{aligned}$

g) $\begin{aligned} \int{{{e}^{3-2x}}\,dx}=-\dfrac{1}{2}{{e}^{3-2x}}+ C\\ \end{aligned}$

$\begin{aligned}h)\, \int{\dfrac{1}{\left( 1+x \right)\left( 1-2x \right)}\,dx}&=\int{\dfrac{1}{3}\left( \dfrac{1}{1+x}+\dfrac{2}{1-2x} \right)\,dx} \\ & =\dfrac{1}{3}\left( \int{\dfrac{1}{1+x}\,dx+\int{\dfrac{2}{1-2x}}\,dx} \right) \\ & =\dfrac{1}{3}\ln \left| 1+x \right|-\dfrac{1}{3}\ln \left| 1-2x \right|+C\\ \end{aligned}$

Ghi nhớ: Một số công tính tính nguyên hàm mở rộng:

$\begin{align} & \int{{{x}^{\frac{m}{n}}}dx}=\dfrac{n}{m+n}{{x}^{\frac{m+n}{n}}}+C;\,\,\,\,\,\int{\sin axdx}=-\dfrac{\operatorname{cosax}}{a}+C; \\ & \int{\cos axdx}=\dfrac{\sin ax}{a}+C;\,\,\,\,\,\,\,\,\,\,\int{{{e}^{ax}}dx}=\dfrac{1}{a}{{e}^{ax}}+C \\ & \int{\frac{1}{a+bx}dx}=\dfrac{1}{b}\ln \left| a+bx \right|+C \\ \end{align}$