Đề thi online Tính nguyên hàm bằng phương pháp đổ...

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

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

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

D \(\int {f\left( x \right)dx = } {{\left( {{x^2} + 3} \right)} \over 2} + C\)

A \(I = {\sin ^4}x+ C\)

B \(I = {1 \over 4}{\sin ^4}x + C\)

C \(I = {1 \over 4}\sin x + C\)

D \(I = {1 \over 2}{\sin ^4}x + C\)

A \(\int {f\left( x \right)dx = {{{{\ln }^2}2x} \over 2} + C} \)

B \(\int {f\left( x \right)dx = {{\ln 2x} \over 2} + C} \)

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

D \(\int {f\left( x \right)dx = {{{{\ln }^2}x} \over 2} + C} \)

A \(\int {f\left( x \right)dx = } {1 \over {8{{\left( {2{x^2} + 3} \right)}^2}}} + C\)

B \(\int {f\left( x \right)dx = } {{ - 1} \over {8{{\left( {2{x^2} + 3} \right)}^2}}} + C\)

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

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

A \(\int {f\left( x \right)dx = {1 \over {2{{\left( {\tan x + 3} \right)}^2}}} + C} \)

B \(\int {f\left( x \right)dx = {{ - 1} \over {2{{\left( {\tan x + 3} \right)}^2}}} + C} \)

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

D \(\int {f\left( x \right)dx = {{ - 1} \over {2{{\left( {\cot x + 3} \right)}^2}}} + C} \)

A \(\int {f\left( x \right)dx = {1 \over {{e^x} - 2}} + C} \)

B \(\int {f\left( x \right)dx = {2 \over {{e^x} - 2}} + C} \)

C \(\int {f\left( x \right)dx = {{ - 2} \over {{e^x} - 2}} + C} \)

D \(\int {f\left( x \right)dx = {{ - 1} \over {{e^x} - 2}} + C} \)

A 1

B 2

C 3

D 4

A \({e^{2\ln x + 3}} + C\)

B \(2{e^{2\ln x + 3}} + C\)

C \({1 \over 2}{e^{2\ln x + 3}} + C\)

D \( - {1 \over 2}{e^{2\ln x + 3}} + C\)

A \(f\left( x \right) = {{x - 1} \over {\sqrt x }}\)

B \(f\left( x \right) = {1 \over {x\ln x}}\)

C \(f\left( x \right) = {{{{\ln }^3}x} \over x}\)

D \(f\left( x \right) = {{ - 2} \over {1 + x}}\)

A Bước 1

B Bước 2

C Bước 3

D Bài toán đúng 

A \(I = \ln \left( {{x^4} + {x^2} + 2} \right) + C\)

B \(I = \ln \left( {{x^4} + {x^2} } \right) + C\)

C \(I = {1 \over {{x^4} + {x^2} + 2}} + C\)

D \(I = {1 \over 2}\ln \left( {{x^4} + {x^2} + 2} \right) + C\)

A \(I = \ln \left| {t - 2} \right| - \ln \left| {t - 1} \right| + C\)

B \(\int {{1 \over {{e^x} + 2{e^{ - x}} - 3}}dx}  = \int {{1 \over {\left( {t - 1} \right)\left( {t - 2} \right)}}dt} \)

C \(I = \int {\left( {{1 \over {t - 1}} - {1 \over {t - 2}}} \right)} dx\)

D \(I = \ln \left| {{{{e^x} - 2} \over {{e^x} - 1}}} \right| + C\)

A \(f\left( t \right) = {t^4}\)

B \(f\left( t \right) = {t^2}\)

C \(f\left( t \right) = {\left( {1 - t} \right)^2}\)

D \(f\left( t \right) = {\left( {1 - t} \right)^3}\)

A \(F\left( x \right) = \ln \left| {{e^x} + 4} \right|\)

B \(F\left( x \right) = \ln \left| {{e^x} + 4} \right| + {e^2}\)

C \(F\left( x \right) = {e^x}\ln \left| {{e^x} + 4} \right|\)

D \(F\left( x \right) = \ln \left| {{e^x} + 4} \right| + \sin \alpha \)

A \(I = {1 \over 2}{\tan ^2}x + \ln \left| {\cos x} \right| + C\)

B \(I = {\tan ^2}x - \ln \left| {\cos x} \right| + C\)

C \(I = {\tan ^2}x + \ln \left| {\cos x} \right| + C\)

D \(I = {1 \over 2}{\tan ^2}x - \ln \left| {\cos x} \right| + C\)

A \(I = {2 \over {15}}\left( {3{x^2} + 4x + 8} \right)\sqrt {1 - x}  + C\)

B \(I = \left( {3{x^2} + 4x + 8} \right)\sqrt {1 - x}  + C\)

C \(I = {{ - 2} \over {15}}\left( {3{x^2} + 4x + 8} \right) + C\)

D \(I = {{ - 2} \over {15}}\left( {3{x^2} + 4x + 8} \right)\sqrt {1 - x}  + C\)

A \(\int {f\left( x \right)dx}  =  - {\cos ^3}x + C\)

B \(\int {f\left( x \right)dx}  =  - {1 \over 3}{\cos ^3}x + C\)

C \(\int {f\left( x \right)dx}  = {1 \over 3}{\sin ^3}x + C\)

D \(\int {f\left( x \right)dx}  =  - {1 \over 3}{\sin ^3}x + C\)

A \( 2a-b>0 \)

B \(2a-b<0\)

C \(3a-b<0\)

D \(a+b=3\)

A \(x=\tan t\)

B \(t=x^2+1\)

C Hai cách đặt trên đều được 

D Hai cách đặt trên đều không được