Nguyên hàm

\(\int{f(x)dx}=\int{\left( {{x}^{3}}+x+1 \right)dx}=\frac{{{x}^{4}}}{4}+\frac{{{x}^{2}}}{2}+x+C\)

\(f(x)=5{{x}^{4}}+2\Rightarrow \int{f(x)dx}=\int{\left( 5{{x}^{4}}+2 \right)dx}={{x}^{5}}+2x+C\)

\(\int{\left( {{x}^{2}}-x+1 \right)dx}=\frac{{{x}^{3}}}{3}-\frac{{{x}^{2}}}{2}+x+C\)

\(\int{\frac{1}{{{x}^{2}}}}dx=\int{{{x}^{-2}}}dx=\frac{{{x}^{-1}}}{-1}+C=-\frac{1}{x}+C\)

\(\int{f(x)dx}=\int{\left( 2\sqrt{x}+3x \right)dx}=2\int{{{x}^{\frac{1}{2}}}dx+3\int{xdx}}=2.\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}+3.\frac{{{x}^{2}}}{2}+C=\frac{4}{3}x\sqrt{x}+\frac{3}{2}{{x}^{2}}+C\)

Ta có \(H = \int {\sqrt[4]{{2x - 1}}{\text{d}}x} = \int {{{\left( {2x - 1} \right)}^{\frac{1}{4}}}{\text{d}}x} \) \( = \dfrac{1}{2}.\dfrac{1}{{\frac{5}{4}}}.{\left( {2x - 1} \right)^{\frac{5}{4}}} + C \) \(= \dfrac{2}{5}{\left( {2x - 1} \right)^{\frac{5}{4}}} + C.\)

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

\(\begin{align}  & F\left( x \right)=\int{f\left( x \right)dx}=\int{{{e}^{2x}}dx}=\frac{{{e}^{2x}}}{2}+C \\  & F\left( 0 \right)=\frac{1}{2}+C=1\Leftrightarrow C=\frac{1}{2} \\  & \Rightarrow F\left( x \right)=\frac{{{e}^{2x}}}{2}+\frac{1}{2} \\ \end{align}\)

$f\left( x \right)=\int\limits_{{}}^{{}}{f'\left( x \right)dx}=\int\limits_{{}}^{{}}{\frac{1}{x+1}dx}$ $=\ln \left| x+1 \right|+C$

$f\left( 0 \right)=2018\Leftrightarrow C=2018$ $\Rightarrow f\left( x \right)=\ln \left| x+1 \right|+2018$ 

$\Rightarrow f\left( 3 \right)-f\left( 1 \right)$ $=\ln 4+2018-\ln 2-2018=\ln 2$

\(F\left( x \right)=\int\limits_{{}}^{{}}{f\left( x \right)dx}=\int\limits_{{}}^{{}}{{{e}^{2x}}dx}=\frac{{{e}^{2x}}}{2}+C\Rightarrow F\left( 0 \right)=\frac{1}{2}+C=\frac{3}{2}\Rightarrow C=1\Rightarrow F\left( x \right)=\frac{{{e}^{2x}}}{2}+1\Rightarrow F\left( \frac{1}{2} \right)=\frac{e}{2}+1\)

Ta có:

+) Đáp án A: \(\left( {2\sqrt {3x - 2}  + C} \right)' = \dfrac{{2.3}}{{2\sqrt {3x - 2} }} = \dfrac{3}{{\sqrt {3x - 2} }} \ne \dfrac{1}{{\sqrt {3x - 2} }} \Rightarrow \) đáp án A sai.

+) Đáp án B: \(\left( {\dfrac{2}{3}\sqrt {3x - 2}  + C} \right)' = \dfrac{{2.3}}{{3.2\sqrt {3x - 2} }} = \dfrac{1}{{\sqrt {3x - 2} }} \Rightarrow \) đáp án B đúng.

Ta có \(F\left( x \right)=\int{f\left( x \right)\,dx}=\int{\left( {{x}^{2}}+2x-3 \right)\,dx}=\frac{{{x}^{3}}}{3}+{{x}^{2}}-3x+C.\)

Mà \(F\left( 0 \right)=4\)

\(\Rightarrow \)\({{\left. \left( \frac{{{x}^{3}}}{3}+{{x}^{2}}-3x+C \right) \right|}_{x\,\,=\,\,0}}=4\Rightarrow C=4.\)

Vậy \(F\left( 1 \right)={{\left. \left( \frac{{{x}^{3}}}{3}+{{x}^{2}}-3x+4 \right) \right|}_{x\,\,=\,\,1}}=\frac{7}{3}.\)

$f(x) = 6x + \sin 3x$ $ \Rightarrow \int {f(x)dx} {\rm{\;}} = \int {(6x + \sin 3x)dx} $ $ = \int {6xdx} {\rm{\;}} + \int {\sin 3xdx}  = 3{x^2} - \dfrac{1}{3}\cos 3x + C$

$ \Rightarrow F(x) = 3{x^2} - \dfrac{1}{3}\cos 3x + C$

$F(0) = \dfrac{2}{3} \Leftrightarrow {3.0^2} - \dfrac{1}{3}.\cos 0 + C = \dfrac{2}{3} \Leftrightarrow C = 1$

$ \Rightarrow F(x) = 3{x^2} - \dfrac{{\cos 3x}}{3} + 1$

Có

\(\begin{array}{l}f\left( x \right) = \int {\left( {2 - 5\sin x} \right)dx} = 2x + 5\cos x + C\\10 = f\left( 0 \right) = 2.0 + 5\cos 0 + C \Rightarrow C = 5\\ \Rightarrow f\left( x \right) = 2x + 5\cos x + 5\end{array}\)

Ta có: \(f\left( 1 \right)=0\Rightarrow a+b=0.\) Do \(f\left( x \right)=a\,x+\frac{b}{{{x}^{2}}}\left( x\ne 0 \right)\Rightarrow F\left( x \right)=\int\limits_{{}}^{{}}{f\left( x \right)dx}=\frac{a\,{{x}^{2}}}{2}-\frac{b}{x}+C\)

Do

\(\begin{align}  & F\left( -1 \right)=1\Rightarrow \frac{a}{2}+b+C=1 \\  & F\left( 1 \right)=4\Rightarrow \frac{a}{2}-b+C=4 \\ \end{align}\)

Suy ra  \(a=\frac{3}{2};b=-\frac{3}{2};c=\frac{7}{4}\Rightarrow F\left( x \right)=\frac{3{{x}^{2}}}{4}+\frac{3}{2x}+\frac{7}{4}\)

Ta có: \(\int {f\left( x \right)dx = \dfrac{{{x^3}}}{3} + 3x + C} \)

Ta có \(f'\left( x \right) = \frac{1}{{{x^2} + x - 2}} \Rightarrow f\left( x \right) = \int {f'\left( x \right)\,{\rm{d}}x}  = \int {\frac{{{\rm{d}}x}}{{{x^2} + x - 2}}}  = \frac{1}{3}\ln \left| {\frac{{x - 1}}{{x + 2}}} \right| + C.\)

Khi đó \(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{3}\ln \frac{{x - 1}}{{x + 2}} + {C_1}\,\,\,\,khi\,\,\,\,x > 1\\\frac{1}{3}\ln \frac{{x - 1}}{{x + 2}} + {C_2}\,\,\,khi\,\,\,\,x <  - \,2\\\frac{1}{3}\ln \frac{{1 - x}}{{x + 2}} + {C_3}\,\,\,khi\,\,\, - \,2 < x < 1\end{array} \right.\)

Mà \(f\left( 0 \right)=\frac{1}{3}\)\( \Rightarrow \,\,{C_3} + \frac{1}{3}\ln \frac{1}{2} = \frac{1}{3} \Rightarrow {C_3} = \frac{1}{3} - \frac{1}{3}\ln \frac{1}{2}\).

Và \(f\left( -\,3 \right)-f\left( 3 \right)=0\Leftrightarrow \frac{1}{3}\ln 4+{{C}_{2}}-\frac{1}{3}\ln \frac{2}{5}-{{C}_{1}}=0\Leftrightarrow {{C}_{2}}-{{C}_{1}}=-\frac{1}{3}\ln 4+\frac{1}{3}\ln \frac{2}{5}.\)

Do đó

\(\begin{array}{l}T = \frac{1}{3}\ln \frac{5}{2} + {C_2} + \frac{1}{3}\ln 2 + {C_3} - \frac{1}{3}\ln \frac{1}{2} - {C_1}\\T = \frac{1}{3}\ln \frac{5}{2} + \frac{1}{3}\ln 2 - \frac{1}{3}\ln \frac{1}{2} - \frac{1}{3}\ln 4 + \frac{1}{3}\ln \frac{2}{5} + \frac{1}{3} - \frac{1}{3}\ln \frac{1}{2}\\T = \frac{1}{3}\ln \left( {\frac{{\frac{5}{2}.2.\frac{2}{5}}}{{\frac{1}{2}.4.\frac{1}{2}}}} \right) + \frac{1}{3}\\T = \frac{1}{3}\ln 2 + \frac{1}{3}.\end{array}\)

Bước 1:

Ta có: $\int {\tan {\mkern 1mu} xdx} {\rm{\;}} = \int {\dfrac{{\sin {\mkern 1mu} xdx}}{{\cos x}}} {\rm{\;}} = {\rm{\;}} - \int {\dfrac{{d(\cos x)}}{{\cos x}}} {\rm{\;}}$$ = {\rm{\;}} - \ln \left| {\cos x} \right| + C$

Bước 2:

$f'(x) = \tan {\mkern 1mu} x,{\mkern 1mu} {\mkern 1mu} \forall x \in \left( { - \dfrac{\pi }{4};\dfrac{{3\pi }}{4}} \right)\backslash \left\{ {\dfrac{\pi }{2}} \right\}$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = {\rm{\;}} - \ln \left( {\cos x} \right) + {C_1},{\mkern 1mu} {\mkern 1mu} x \in \left( { - \dfrac{\pi }{4};\dfrac{\pi }{2}} \right)}\\{f(x) = {\rm{\;}} - \ln \left( { - \cos x} \right) + {C_2},{\mkern 1mu} {\mkern 1mu} x \in \left( {\dfrac{\pi }{2};\dfrac{{3\pi }}{4}} \right)}\end{array}} \right.$

Bước 3:

Mà $f(0) = 0,{\mkern 1mu} {\mkern 1mu} f(\pi ) = 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{ - \ln 1 + {C_1} = 0}\\{ - \ln 1 + {C_2} = 1}\end{array}} \right.$$ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{C_1} = 0}\\{{C_2} = 1}\end{array}} \right.$

$ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = {\rm{\;}} - \ln \left( {\cos x} \right),{\mkern 1mu} {\mkern 1mu} x \in \left( { - \dfrac{\pi }{4};\dfrac{\pi }{2}} \right)}\\{f(x) = {\rm{\;}} - \ln \left( { - \cos x} \right) + 1,{\mkern 1mu} {\mkern 1mu} x \in \left( {\dfrac{\pi }{2};\dfrac{{3\pi }}{4}} \right)}\end{array}} \right.$

Bước 4:

$ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f\left( {\dfrac{{2\pi }}{3}} \right) = {\rm{\;}} - \ln \left( { - \cos \dfrac{{2\pi }}{3}} \right) + 1 = \ln 2 + 1}\\{f\left( {\dfrac{\pi }{4}} \right) = {\rm{\;}} - \ln \left( {\cos \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\ln 2}\end{array}} \right.$$ \Rightarrow \dfrac{{f\left( {\dfrac{{2\pi }}{3}} \right)}}{{f\left( {\dfrac{\pi }{4}} \right)}} = \dfrac{{2\left( {\ln 2 + 1} \right)}}{{\ln 2}} = 2\left( {1 + {{\log }_2}e} \right)$

Ta có

+ \(I = \int {\dfrac{1}{{2x}}dx}  = \dfrac{1}{2}\int {\dfrac{1}{x}dx}  = \dfrac{1}{2}\ln \left| x \right| + C\) nên An sai

+ \(I = \int {\dfrac{1}{{2x}}dx}  = \dfrac{1}{2}\int {\dfrac{2}{{2x}}dx}  = \dfrac{1}{2}\int {\dfrac{{d\left( {2x} \right)}}{{2x}} = \dfrac{1}{2}\ln \left| {2x} \right| + C} \) nên Bình sai

Ta thấy cả hai bạn An và Bình đều làm sai vì thiếu dấu giá trị tuyệt đối.

\(\int {{3^{x + 1}}dx}  = \dfrac{{{3^{x + 1}}}}{{\ln 3}} + C\)