150 câu trắc nghiệm Nguyên hàm - Tích phân cơ bản (P3)

Chọn B

$f\left(x\right)=\int f\text{'}\left(x\right)dx=3x+5\cos x+C$

Do $f\left(0\right)=10\iff C=5$

Vậy $f\left(x\right)=3x+5\cos x+5\Rightarrow f\left(\mathrm{\pi }\right)=3\mathrm{\pi }$

Chọn C

Đặt $u=x, dv=\cos xdx$

Suy ra $du=dx, v=\sin x$

Do đó $I=x\sin x-\int \sin xdx=x\sin x+\cos x+C$

Chọn C

$$\begin{gathered} F\left(x\right)=\int {\sin }^2\frac{x}{2}dx=\frac{1}{2}\int \left(1-\cos x\right)dx \\ =\frac{x}{2}-\frac{1}{2}\sin x+C \end{gathered}$$

$F\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{4}\iff \frac{\mathrm{\pi }}{4}-\frac{1}{2}\sin \frac{\mathrm{\pi }}{2}+C=\frac{\mathrm{\pi }}{4}\iff C=\frac{1}{2}$

Vậy $F\left(x\right)=\frac{x}{2}-\frac{\sin x}{2}+\frac{1}{2}$

Chọn B

$$\begin{gathered} \int f\left(x\right)dx=\int \left(e^x\ln 2+\frac{1}{{\sin }^{2}x}\right)dx \\ =e^x\ln 2-\mathrm{co}tx+C \end{gathered}$$

Chọn C

$\int f\left(x\right)dx=\int x.e^{x}dx$

Đặt $u=x\Rightarrow du=dx$ và $dv=e^{x}dx\Rightarrow v=e^x$

Vậy $$\begin{gathered} \int f\left(x\right)dx=x.e^x-\int e^{x}dx \\ =x.e^x-e^x+C \end{gathered}$$

Chọn D

$\int \frac{xdx}{x^2+4}=\frac{1}{2}\int \frac{d(x^2+4)}{x^2+4}=\frac{1}{2}\ln \left(x^2+4\right)+C$

$$\begin{gathered} \int e^{2\cos x}\sin xdx=-\int e^{2\cos x}d\left(\cos x\right) \\ =-\frac{1}{2}{e}^{2\cos x}+C \end{gathered}$$

$$\begin{gathered} \int cotxdx=\int \frac{\cos x}{\sin x}dx=\int \frac{d\left(\sin x\right)}{\sin x} \\ =\ln \left|\sin x\right|+C \\ \ne -\frac{1}{{\sin }^{2}x}+C \end{gathered}$$

Vậy chỉ có (I) và (III) đúng.

Chọn B

Đặt $\left\{\begin{array}{l}u=\ln x\\ dv=xdx\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x}dx\\ v=\frac{x^2}{2}\end{array}\right.$

$$\begin{gathered} \Rightarrow \int x\ln xdx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}.\frac{1}{x}dx \\ =\frac{x^2}{2}\ln x-\frac{1}{2}\int xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C \end{gathered}$$

Chọn B

Đặt $\left\{\begin{array}{l}u=1-x\\ dv=\cos xdx\end{array}\Rightarrow \left\{\begin{array}{l}du=-dx\\ v=\sin x\end{array}\right.\right.$

$$\begin{gathered} \Rightarrow \int \left(1-x\right)\cos xdx=\left(1-x\right)\sin x+\int \sin xdx \\ =\left(1-x\right)\sin x-\cos x+C \end{gathered}$$

Chọn C

Đặt $\left\{\begin{array}{l}u=1-2x\\ dv=e^{x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=-2dx\\ v=e^x\end{array}\right.\right.$

$$\begin{gathered} \Rightarrow \int (1-2x)e^{x}dx=(1-2x)e^x+\int 2e^{x}dx \\ =\left(1-2x\right)e^x+2e^x+C=e^x\left(3-2x\right)+C \end{gathered}$$

Chọn D

Đặt $\left\{\begin{array}{l}u=\ln x\\ dv=\sqrt{x}dx\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x}dx\\ v=\frac{2}{3}{x}^{\frac{3}{2}}\end{array}\right.$

$$\begin{gathered} \Rightarrow \int \sqrt{x}\ln xdx=\frac{2}{3}{x}^{\frac{3}{2}}\ln x-\frac{2}{3}\int x^{\frac{3}{2}}.\frac{dx}{x} \\ =\frac{2}{3}{x}^{\frac{3}{2}}\ln x-\frac{2}{3}\int x^{\frac{1}{2}}dx \\ =\frac{2}{3}{x}^{\frac{3}{2}}\ln x-\frac{2}{3}.\frac{2}{3}{x}^{\frac{3}{2}}+C \\ =\frac{2}{3}{x}^{\frac{3}{2}}\ln x-\frac{4}{9}{x}^{\frac{3}{2}}+C \end{gathered}$$

Chọn A

Đặt $\left\{\begin{array}{l}u=x\\ dv=\frac{1}{{\sin }^{2}x}dx\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}du=dx\\ v=-cotx\end{array}\right.$

$$\begin{gathered} \Rightarrow \int \frac{x}{{\sin }^{2}x}dx=-xcotx+\int \frac{\cos x}{\sin x}dx \\ =-x.cotx+\ln \left|\sin x\right|+C \end{gathered}$$

Chọn B

Đặt $\left\{\begin{array}{l}u=2x+3\\ dv=e^{-x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=2dx\\ v=-e^{-x}\end{array}\right.\right.$

$$\begin{gathered} \Rightarrow \int (2x+3)e^{-x}dx \\ =-e^{-x}(2x+3)-\int \left(-e^{-x}\right).2dx \\ =-e^{-x}(2x+3)+\int 2e^{-x}dx \\ =-e^{-x}(2x+3)-2e^{-x}+C \\ =-e^{-x}(2x+1)+C \end{gathered}$$

Chọn A

$$\begin{gathered} \int \frac{2x^2+2x+3}{2x+1}dx=\int \left(\frac{2x+1}{2}+\frac{5}{2\left(2x+1\right)}\right)dx \\ =\frac{1}{8}(2x+1{)}^2+\frac{5}{4}\ln \left|2x+1\right|+C \end{gathered}$$

Chọn B

$\int \frac{1}{x(\ln x+1)}dx=\int \frac{d(\ln x+1)}{\ln {\displaystyle x}{\displaystyle +}{\displaystyle 1}}=\ln \left|\ln x+1\right|+C$

Chọn A

$$\begin{gathered} \int \frac{e^{2x}}{e^x+1}dx=\int \left(e^x-\frac{e^x}{e^x+1}\right)dx \\ =e^x-\int \frac{d(e^x+1)}{e^x+1}=e^x-\ln (e^x+1)+C \end{gathered}$$

Chọn B

Đặt $t=1+\sqrt{x}\Rightarrow x={(t-1)}^2$$\Rightarrow dx=2\left(t-1\right)dt$

Khi đó 

$$\begin{gathered} \int \frac{1}{1+\sqrt{x}}dx=\int \frac{2(t-1)}{t}dt=2\int \left(1-\frac{1}{t}\right)dt \\ =2(t-\ln \left|t\right|)+C_1 \\ =2\left(\sqrt{x}+1-\ln \left|1+\sqrt{x}\right|\right)+C_1 \\ =2\sqrt{x}-2\ln \left(1+\sqrt{x}\right)+C \end{gathered}$$

(Với $C=2+C_1$ và $1+\sqrt{x}>0$)

Chọn A

$$\begin{gathered} \int \frac{2x-1}{\sqrt{1-x}}dx=-\int \left(-2\sqrt{1-x}+\frac{1}{\sqrt{1-x}}\right)d(1-x) \\ =\frac{4}{3}{\left(1-x\right)}^{\frac{3}{2}}-2{\left(1-x\right)}^{\frac{1}{2}}+C=-\frac{2}{3}(2x+1)\sqrt{1-x}+C \end{gathered}$$

Chọn A

Đặt $t=\sqrt{4-x^2}\Rightarrow x^2=4-t^2$$\Rightarrow xdx=-tdt$

Khi đó $$\begin{gathered} \int \frac{x^3}{\sqrt{4-x^2}}dx=\int \frac{(4-t^2)(-tdt)}{t} \\ =\int (t^2-4)dt=\frac{t^3}{3}-4t+C \\ =\frac{{\displaystyle {\left(\sqrt{4-x^2}\right)}^3}}{3}-4\sqrt{4-x^2}+C \\ =-\frac{1}{3}\left(x^2+8\right)\sqrt{4-x^2}+C \end{gathered}$$

Chọn B

$$\begin{gathered} \int \frac{1}{\sqrt{x}{\cos }^2\sqrt{x}}dx \\ =2\int \frac{1}{{\displaystyle {\cos }^2\sqrt{x}}}d\left(\sqrt{x}\right)=2\tan \sqrt{x}+C \end{gathered}$$

Chọn A

$$\begin{gathered} \int \frac{6x^2-12x}{x^3-3x^2+6}dx=2\int \frac{1}{x^3-3x^2+6}d\left(x^3-3x^2+6\right) \\ =2\ln \left|x^3-3x^2+6\right|+C \end{gathered}$$

Chọn C

$$\begin{gathered} \int {(5-9x)}^{12}dx=-\frac{1}{9}\int {\left(5-9x\right)}^{12}d\left(5-9x\right) \\ =-\frac{{\left(5-9x\right)}^{13}}{117}+C \end{gathered}$$

Chọn B

$$\begin{gathered} \int \frac{1}{(\cos x+\sin x{)}^2}dx \\ =\frac{1}{2}\int \frac{1}{{\displaystyle {\sin }^2\left(x+\frac{\mathrm{\pi }}{4}\right)}}dx \\ =\frac{{\displaystyle 1}}{{\displaystyle 2}}\int \frac{{\displaystyle 1}}{{\displaystyle {\sin }^2\left(x+\frac{\mathrm{\pi }}{4}\right)}}d\left(x+\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right) \\ =-\frac{1}{2}cot\left(x+\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right)+C \end{gathered}$$

Chọn D

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

Chọn A

Đặt $\left\{\begin{array}{l}u=x\\ dv=2^{x}dx\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}du=dx\\ v=\frac{2^x}{\ln 2}\end{array}\right.$

Ta có $\int x.2^{x}dx=\frac{x.2^x}{\ln 2}-\int \frac{2^x}{\ln 2}dx=$$\frac{x.2^x}{\ln 2}-\frac{2^x}{{\displaystyle {\ln }^2}{\displaystyle 2}}+C$

Chọn D

Đặt $\left\{\begin{array}{l}u=\ln x\\ dv=dx\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x}dx\\ v=x\end{array}\right.$

Ta có $\int \ln xdx=x\ln x-\int dx=x\ln x-x+C$

Chọn C

Đặt 

$$\begin{gathered} \left\{\begin{array}{l}u=\ln (x-1)\\ dv=2xdx\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}du=\frac{1}{x-1}dx\\ v=x^2-1\end{array}\right. \end{gathered}$$

Ta có 

$$\begin{gathered} \int 2x\ln (x-1)dx \\ =(x^2-1)\ln (x-1)-\int (x+1)dx \\ =(x^2-1)\ln (x-1)-\frac{x^2}{2}-x+C \end{gathered}$$

Chọn B

Ta có ${\int }_{-1}^0\left(x+1+\frac{2}{x-1}\right)dx=\left(\frac{x^2}{2}+x+2\ln \left|x-1\right|\right)\overline{)\begin{array}{c}0\\ -1\end{array}}$$=\frac{1}{2}-2\ln 2=a+b\ln 2$$\Rightarrow \left\{\begin{array}{l}a=\frac{1}{2}\\ b=-2\end{array}\right.$

Vậy $a+b=\frac{1}{2}-2=\frac{-3}{2}$

Chọn B

Đặt $\left\{\begin{array}{l}u=\ln (x+1)\\ dv=dx\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x+1}dx\\ v=x+1\end{array}\right.$

Khi đó 

$$\begin{gathered} {\int }_1^2\ln (x+1)dx \\ =(x+1)\ln (x+1)\overline{){}_1^2}-{\int }_1^{2}dx \\ =3\ln 3-2\ln 2-1 \end{gathered}$$

Vậy a=3; b=-2; c=-1$\Rightarrow S=a+b+c=0$

Chọn C

Ta có 

$$\begin{gathered} I=4{\int }_1^{e}x(1+\ln x)dx \\ =2{\int }_1^e(1+\ln x)d\left(x^2\right) \\ =2\left[\left(1+\ln x\right)x^2{|}_1^e-{\int }_1^e{x}^2.\frac{1}{x}dx\right] \\ =2\left(2e^2-1-\frac{e^2}{2}+\frac{1}{2}\right)=3e^2-1 \end{gathered}$$

Nên a=3; b=-1 nên M=5.

Chọn A

Cho $x-1=0\iff x=1$(thỏa mãn)

Ta có bảng xét dấu:

Khi đó 

$$\begin{gathered} I=-{\int }_0^1(x-1)dx+{\int }_1^2(x-1)dx \\ =-\left(\frac{x^2}{2}-x\right){|}_0^1+\left(\frac{x^2}{2}-x\right){|}_1^2=1 \end{gathered}$$