Trắc nghiệm Nguyên hàm lớp 12 (có đáp án - phần 2)
30 câu hỏi
\[\int {f(x)dx = } F(x) + C\].
\({\left( {\int {f(x)dx} } \right)^\prime } = f(x)\).
\({\left( {\int {f(x)dx} } \right)^\prime } = f'(x)\).
\({\left( {\int {f(x)dx} } \right)^\prime } = F'(x)\).
Chọn C
Ta có \[\int {f(x)dx = } F(x) + C \Leftrightarrow F'\left( x \right) = f\left( x \right)\] nên phương án A, B, D đúng.
\[{x^2} + C\].
\[{x^2} + 6x + C\].
\[2{x^2} + C\].
\[2{x^2} + 6x + C\].
Chọn B
\(\int {\left( {2x + 6} \right)dx = {x^2} + 6x + C} \)
\(2x + C\).
\(\frac{1}{3}{x^3} + C\).
\({x^3} + C\).
\(3{x^3} + C\)
Chọn B
Ta có \(\int {{x^2}dx} = \frac{1}{3}{x^3} + C\).
\({x^3} + C\)
\(\frac{{{x^3}}}{3} + x + C\)
\(6x + C\)
\({x^3} + x + C\)
Chọn D
\(\int {\left( {3{x^2} + 1} \right)dx = {x^3} + x + C.} \)
\(\frac{1}{4}{x^4} + \frac{1}{2}{x^2} + C\)
\(3{x^2} + 1 + C\)
\({x^3} + x + C\)
\({x^4} + {x^2} + C\)
Chọn A
\(\int {\left( {{x^3} + {x^2}} \right){\rm{d}}x} \)\( = \frac{1}{4}{x^4} + \frac{1}{2}{x^2} + C\).
\[\frac{1}{5}{x^5} + \frac{1}{3}{x^3} + C\]
\[{x^4} + {x^2} + C\]
\[{x^5} + {x^3} + C\].
\[4{x^3} + 2x + C\]
Chọn A
\[\int {f\left( x \right)dx = } \]\[\int {\left( {{x^4} + {x^2}} \right)dx} \]\[ = \frac{1}{5}{x^5} + \frac{1}{3}{x^3} + C\].
\(\frac{{{x^{2023}}}}{{2023}} + 1\).
\(\frac{{{x^{2023}}}}{{2023}}\).
\(y = 2022{x^{2021}}\).
\(\frac{{{x^{2023}}}}{{2023}} - 1\).
Chọn C
Ta có: \(\int {{x^{2022}}{\rm{d}}} x = \frac{{{x^{2023}}}}{{2023}} + C,{\rm{ }}C\)là hằng số.
Nên các phương án A, B, D đều là nguyên hàm của hàm số \(y = {x^{2022}}\).
\[\frac{1}{{12}}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} + C\].
\(\frac{1}{9}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} - 2024x + C\).
\(\frac{1}{{12}}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} - 2024x + C\).
\(\frac{1}{9}{x^4} + \frac{2}{3}{x^3} - \frac{{{x^2}}}{2} - 2024x + C\).
Chọn C
Sử dụng công thức \(\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \) ta được:
\(\int {\left( {\frac{1}{3}{x^3} - 2{x^2} + x - 2024} \right)dx = } \frac{1}{3}.\frac{{{x^4}}}{4} - 2.\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} - 2024x + C = \frac{1}{{12}}{x^4} - \frac{2}{3}{x^3} + \frac{1}{2}{x^2} - 2024x + C.\)
\[F\left( x \right)\, = \,\frac{{{x^4}}}{4}\, - \,6{x^3}\, + \,\frac{{11}}{2}{x^2} - \,6x\, + \,C\].B. \[F\left( x \right)\, = \,{x^4}\, + \,6{x^3}\, + \,11{x^2} + \,6x\, + \,C\].
\[F\left( x \right)\, = \,\frac{{{x^4}}}{4}\, + 2{x^3}\, + \,\frac{{11}}{2}{x^2} + \,6x\, + \,C\].D. \[F\left( x \right)\, = \,{x^3}\, + \,6{x^2}\, + \,11{x^2} + \,6x\, + \,C\].
Chọn C
Ta có: \[f\left( x \right)\, = \,\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = \,{x^3} + \,6{x^2}\, + \,11x\, + \,6\]
\[ \Rightarrow \,\,F\left( x \right)\, = \,\int {\left( {{x^3} + \,6{x^2}\, + \,11x\, + \,6} \right)} dx\, = \,\,\frac{{{x^4}}}{4}\, + 2{x^3}\, + \,\frac{{11}}{2}{x^2} + \,6x\, + \,C\].
Học thông minh hơn mỗi ngày
Xem toàn bộ 30 câu hỏi và nhiều hơn nữa với tài khoản miễn phí
\(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^3}}}{3} + \frac{1}{x} + C\).
\(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^3}}}{3} - \frac{2}{x} + C\).
\(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^3}}}{3} - \frac{1}{x} + C\).
\(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^3}}}{3} + \frac{2}{x} + C\).
Chọn A
Ta có \(\int {\left( {{x^2} + \frac{2}{{{x^2}}}} \right){\rm{d}}x} = \frac{{{x^3}}}{3} - \frac{2}{x} + C\).
\(\frac{4}{{15}}x\sqrt[{15}]{{{x^7}}} + C\).
\(\frac{8}{{15}}x\sqrt[{15}]{{{x^7}}} + C\).
\(\frac{8}{{15}}x\sqrt[{15}]{x} + C\).
\(\frac{4}{{15}}x\sqrt[{15}]{x} + C\).
Chọn B
\(\int {\sqrt {x\sqrt {x\sqrt x } } dx} = \int {\sqrt {x\sqrt {x.{x^{\frac{1}{2}}}} } dx} = \int {\sqrt {x.{x^{\frac{3}{4}}}} dx} = \int {{x^{\frac{7}{8}}}dx} = \frac{{{x^{\frac{7}{8} + 1}}}}{{\frac{7}{8} + 1}} + C = \frac{8}{{15}}x\sqrt[{15}]{{{x^7}}} + C\)
\[x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].
\[\frac{4}{5}x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].
\[x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].
\[\frac{4}{5}x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].
Chọn B
\[\int {\frac{{\sqrt x - 2\sqrt[3]{{{x^2}}} + 1}}{{\sqrt[4]{x}}}dx} = \int {\frac{{{x^{\frac{1}{2}}} - 2{x^{\frac{2}{3}}} + 1}}{{{x^{\frac{1}{4}}}}}dx = } \int {\left( {\frac{{{x^{\frac{1}{2}}}}}{{{x^{\frac{1}{4}}}}} - 2\frac{{{x^{\frac{2}{3}}}}}{{{x^{\frac{1}{4}}}}} + \frac{1}{{{x^{\frac{1}{4}}}}}} \right)dx} \]
\[ = \int {\left( {{x^{\frac{1}{4}}} - 2{x^{\frac{5}{{12}}}} + {x^{ - \frac{1}{4}}}} \right)dx = \frac{4}{5}} x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\]
\(\int f (x)dx = 2x + C\).
\(\int f (x)dx = {x^2} + 4x + C\).
\(\int f (x)dx = \frac{{{x^3}}}{3} + 4x + C\).
\(\int f (x)dx = {x^3} + 4x + C\).
Ta có: \(f(x) = {x^2} + 4 \Rightarrow \int f (x)dx = \frac{{{x^3}}}{3} + 4x + C\)
\(\int {f\left( x \right)} dx = \frac{3}{2}{x^{\frac{1}{2}}} + C\).
\(\int {f\left( x \right)} dx = \int {\sqrt {{x^3}} } dx\).
\(\int {f\left( x \right)} dx = \frac{2}{5}{x^{\frac{5}{2}}} + C\).
\(\int {f\left( x \right)} dx = \frac{2}{3}{x^{\frac{1}{2}}} + C\).
Ta có:\(\int {f\left( x \right)} dx = \int {{x^{\frac{3}{2}}}} dx = \frac{2}{5}{x^{\frac{5}{2}}} + C\).
\(\int {f\left( x \right){\rm{d}}x = } \frac{{{x^3}}}{3} - \frac{1}{x} + C\).
\(\int {f\left( x \right){\rm{d}}x = } \frac{{{x^3}}}{3} + \frac{2}{x} + C\).
\[\int {f\left( x \right){\rm{d}}x = } \int {\left( {{x^2} + \frac{2}{{{x^2}}}} \right)} {\rm{d}}x\].
\(\int {f\left( x \right){\rm{d}}x = } \frac{{{x^3}}}{3} - \frac{2}{x} + C\).
A. Ta có: \[\int {f\left( x \right){\rm{d}}x = } \int {\frac{{{x^4} + 2}}{{{x^2}}}} {\rm{d}}x = \int {\left( {{x^2} + \frac{2}{{{x^2}}}} \right)} {\rm{d}}x = \frac{{{x^3}}}{3} - \frac{2}{x} + C\].
\(f\left( x \right) = 6{x^2} - 2\)
\(f\left( x \right) = \frac{1}{2}{x^4} - {x^2} + x\)
\(f\left( x \right) = \frac{1}{2}{x^4} - {x^2} + x + C\).
\(f\left( x \right) = 6{x^2} - 2 + C\).
Chọn A
Ta có: \(f(x) = F'(x) = {\left( {2{x^3} - 2x + 1} \right)^\prime } = 6{x^2} - 2\).
\(\int {F'\left( x \right)dx = F\left( x \right) + C} \).
\(\int {F\left( x \right)dx = F'\left( x \right) + C} \).
\(\int {F\left( x \right)dx = F\left( x \right) + C} \).
\(\int {F'\left( x \right)dx = F'\left( x \right) + C} \).
Chọn A
\[\int {{e^{ - 3x}}dx} = {e^{ - 3x}} + C\].
\[\int {{e^{ - 3x}}dx} = - \frac{1}{3}{e^{ - 3x}} + C\].
\[\int {{e^{ - 3x}}dx} = \frac{1}{3}{e^{ - 3x}} + C\].
\[\int {{e^{ - 3x}}dx} = - \frac{1}{3}{e^{ - 3x}}\].
Chọn B
\[\frac{{{{\left( {2x} \right)}^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}} + C\].
\[\frac{{{2^{\sqrt 2 }}{x^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}} + C\].
\[\frac{{{{\left( {2x} \right)}^{\sqrt 2 }}}}{{\ln \left( {2x} \right)}} + C\].
\[{\left( {2x} \right)^{\sqrt 2 }} + C\].
Chọn B
Đặt \[t = 2x \Rightarrow {\rm{d}}t = 2{\rm{d}}x \Rightarrow {\rm{d}}x = \frac{1}{2}{\rm{d}}t\].
Ta có \[\int {{{\left( {2x} \right)}^{\sqrt 2 }}{\rm{d}}x} = \frac{1}{2}\int {{t^{\sqrt 2 }}dt} = \frac{1}{2}\frac{{{t^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}} + C\].
Thay \[t = 2x\] ta có \[\int {{{\left( {2x} \right)}^{\sqrt 2 }}{\rm{d}}x} = \frac{1}{2}\frac{{{{\left( {2x} \right)}^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}} + C = \frac{{{2^{\sqrt 2 }}{x^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}} + C\].
\[x - \cos x + C\].
\[{\left( { - \cos \frac{x}{2} + \sin \frac{x}{2}} \right)^2} + C\].
\[\frac{1}{3}{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)^3} + C\].
\[x + \cos x + C\].
Chọn A
Ta có: \[{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)^2} = {\sin ^2}\frac{x}{2} + 2\sin \frac{x}{2}.\cos \frac{x}{2} + {\cos ^2}\frac{x}{2} = 1 + \sin x\].
Khi đó \[\int {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}{\rm{d}}x} = \int {\left( {1 + \sin x} \right){\rm{d}}x} = \int {{\rm{d}}x} + \int {\sin x{\rm{d}}x = x - \cos x + C} \].
\({f_2}\left( x \right) = \frac{1}{{{{\sin }^2}x}}\).
\({f_1}\left( x \right) = - \frac{1}{{{{\cos }^2}x}}\).
\({f_4}\left( x \right) = \frac{1}{{{{\cos }^2}x}}\).
\({f_3}\left( x \right) = - \frac{1}{{{{\sin }^2}x}}\).
Chọn D
Có \(\int {\frac{1}{{{{\sin }^2}x}}dx} = - \cot x + C\) suy ra \(F\left( x \right) = \cot x\) trên khoảng \(\left( {0;\frac{\pi }{2}} \right)\) là một nguyên hàm của hàm số \({f_3}\left( x \right) = - \frac{1}{{{{\sin }^2}x}}\).
\(\int {f\left( x \right){\rm{d}}x} = x - \cos x + C\).
\(\int {f\left( x \right){\rm{d}}x} = x + \sin x + C\).
\(\int {f\left( x \right){\rm{d}}x} = x + \cos x + C\).
\(\int {f\left( x \right){\rm{d}}x} = \cos x + C\).
Chọn A
Ta có \[\int {f\left( x \right){\rm{d}}x = \int {\left( {1 + \sin x} \right){\rm{d}}x} = \int {1{\rm{d}}x} + \int {\sin x{\rm{d}}x} = x - \cos x + C} \].
A. \(F\left( x \right) = 2\cos \frac{x}{2} + C\)
B. \(F\left( x \right) = \frac{1}{2}\left( {1 + \sin x} \right) + C\)
C. \(F\left( x \right) = 2\sin \frac{x}{2} + C\)
D. \(F\left( x \right) = \frac{1}{2}\left( {1 - \sin x} \right) + C\)
Chọn B
B. Ta có:\(f\left( x \right) = {\cos ^2}\frac{x}{2} \Rightarrow F\left( x \right) = \int {{{\cos }^2}\frac{x}{2}dx} = \int {\frac{{1 + \cos x}}{2}dx} = \frac{1}{2}\int {\left( {1 + \cos x} \right)dx} = \frac{1}{2}\left( {1 + \sin x} \right) + C\)
\(\int {f\left( x \right){\rm{d}}x} = x + \tan x + C\).
\(\int {f\left( x \right){\rm{d}}x} = x + \cot x + C\).
\(\int {f\left( x \right){\rm{d}}x} = x - \tan x + C\).
\(\int {f\left( x \right){\rm{d}}x} = x - \cot x + C\).
Chọn C
\(\int {f\left( x \right){\rm{d}}x} = \int {\left( {1 - \frac{1}{{{{\cos }^2}x}}} \right){\rm{d}}x} = x - \tan x + C\).
\(\sin x + 3{x^2} + C\).
\( - \sin x + 3{x^2} + C\).
\(\sin x + 6{x^2} + C\).
\( - \sin x + C\).
Chọn A
C. Ta có \(\int {f\left( x \right){\rm{d}}x = \int {\left( {\cos x + 6x} \right){\rm{d}}x = \sin x + 3{x^2} + C} } \).
\(\int {\left( {2\sin x + 3x} \right)dx = - 2\cos x + \frac{3}{2}{x^2} + C} \)
\(\int {\left( {2\sin x + 3x} \right)dx = 2\cos x + 3{x^2} + C} \)
\(\int {\left( {2\sin x + 3x} \right)dx = {{\sin }^2}x + \frac{3}{2}x + C} \)
\(\int {\left( {2\sin x + 3x} \right)dx = \sin 2x + \frac{3}{2}{x^2} + C} \)
Chọn A
\(\int {\left( {2\sin x + 3x} \right)dx = - 2\cos x + \frac{3}{2}{x^2} + C} \)
\(\frac{{{x^2}}}{2} + \sin x + C\).
\(\frac{{{x^2}}}{2} - \cos x + C\).
\(\frac{{{x^2}}}{2} - \sin x + C\).
\(\frac{{{x^2}}}{2} + \cos x + C\).
Chọn D
Ta có \(\int {\left( {x - \sin x} \right){\rm{d}}x\,{\rm{ = }}\,} \frac{{{x^2}}}{2} + \cos x + C\).
\[{x^3} + \cos x + C\].
\[6x + \cos x + C\].
\[{x^3} - \cos x + C\].
\[6x - \cos x + C\].
Chọn C
Ta có \[\int {\left( {3{x^2} + \sin x} \right){\rm{d}}x} = {x^3} - \cos x + C\].
\(\ln x - \cos x + C\).
\( - \frac{1}{{{x^2}}} - \cos x + C\).
\(\ln \left| x \right| + \cos x + C\).
\(\ln \left| x \right| - \cos x + C\).
Chọn D
Ta có \(\int {f\left( x \right){\rm{d}}x} = \int {\left( {\frac{1}{x} + \sin x} \right){\rm{d}}x} = \int {\frac{1}{x}{\rm{d}}x} + \int {\sin x{\rm{d}}x} = \ln \left| x \right| - \cos x + C\).
\(f\left( x \right) = - \sin x\).
\(f\left( x \right) = - \cos x\).
\(f\left( x \right) = \sin x\).
\(f\left( x \right) = \cos x\).
Áp dụng công thức \(\smallint {\rm{sin}}x{\rm{\;d}}x = - {\rm{cos}}x + C\). Suy ra \(f\left( x \right) = {\rm{sin}}x\).
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