150 câu trắc nghiệm Nguyên hàm - Tích phân cơ bản (P2)

Chọn D

Ta có: $\frac{1}{x(x+3)}=\frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}\right)$

Nên $$\begin{gathered} \int f\left(x\right)dx=\int \frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}\right)dx \\ =\frac{1}{3}\ln \left|\frac{x}{x+3}\right|+C \end{gathered}$$

Chọn A

$f\left(x\right)=\frac{1}{x^2+x-2}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$

Nên $$\begin{gathered} \int f\left(x\right)dx=\frac{1}{3}\left(\ln \left|x-1\right|-\ln \left|x+2\right|\right)+C \\ =F\left(x\right)=\frac{1}{3}\ln \left|\frac{x-1}{x+2}\right|+C \end{gathered}$$

Chọn A

$$\begin{gathered} f\left(x\right)={\left(\frac{1-x}{x}\right)}^2 \\ =\frac{1-2x+x^2}{x^2} \\ =\frac{1}{x^2}-\frac{2}{x}+1 \end{gathered}$$

Nên $\int f\left(x\right)dx=-\frac{1}{x}-2\ln \left|x\right|+x+C$

Chọn D

Đặt $t=\sqrt{8-x^2}$$\Rightarrow t^2=8-x^2$$\Rightarrow -tdt=xdx$

$$\begin{gathered} \int \frac{x}{\sqrt{8-x^2}}dx=-\int \frac{tdt}{t} \\ =-t+C=-\sqrt{8-x^2}+C \end{gathered}$$

Vì $F\left(2\right)=0$ nên $-\sqrt{8-4}+C=0$ suy ra C = 2.

Ta có phương trình $-\sqrt{8-x^2}+2=x$

$\begin{array}{l}\sqrt{8-x^2}-2-x\\ \Rightarrow 8-x^2={\left(2-x\right)}^2\\ \iff x^2-2x-2=0\end{array}$

$\iff \left[\begin{array}{c}x=1-\sqrt{3}\\ x=1+\sqrt{3}\end{array}\right.$

Thử lại ta thấy $x=1-\sqrt{3}$ thỏa mãn 

Vậy nghiệm của phương trình là $x=1-\sqrt{3}$.

Chọn C

$\int \frac{1}{x-1}dx=\ln \left|x-1\right|+C$

Vì $F\left(2\right)=1$ nên C = 1.

Vậy $F\left(x\right)=\ln \left|x-1\right|+1$, thay x = 3 ta được F(3) = ln2 + 1

Chọn A

Đặt $t=\sqrt{{\ln }^2 x+1}$$\Rightarrow t^2={\ln }^{2}x+1$$\Rightarrow tdt=\frac{\ln x}{x}dx$

$$\begin{gathered} \int \sqrt{{\ln }^{2}x+1}.\frac{\ln x}{x}dx=\int t^{2}dt \\ =\frac{t^3}{3}+C=\frac{{\left(\sqrt{{\ln }^{2}x+1}\right)}^3}{3}+C \end{gathered}$$

Vì $F\left(1\right)=\frac{1}{3}$ nên $C=0$

Vậy $F^2\left(e\right)=\frac{8}{9}$

Chọn D

Đặt $t=\sqrt[3]{3-2x}$$\Rightarrow t^3=3-2x\iff x=\frac{3-t^3}{2}$$\Rightarrow dx=-\frac{3}{2}{t}^{2}dt$

$$\begin{gathered} \Rightarrow I=-\frac{3}{2}\int \left(\frac{3-t^3}{2}+1\right)t.t^{2}dt \\ =-\frac{3}{4}\int (5t^3-t^6)dt=-\frac{3}{4}\left(\frac{5t^4}{4}-\frac{t^7}{7}\right)+C \\ =\frac{3}{4}\left(\frac{\sqrt[3]{{\left(3-2x\right)}^7}}{7}-\frac{5\sqrt[3]{{\left(3-2x\right)}^4}}{4}\right)+C \end{gathered}$$

Chọn B

Đặt $t=\sqrt[3]{2x+2}$$\Rightarrow t^3=2x+2$$\Rightarrow x=\frac{t^3-2}{2}$$\Rightarrow dx=\frac{3}{2}{t}^{2}dt$

Suy ra 

$$\begin{gathered} J=\int \frac{{\displaystyle \frac{t^3-2}{2}}.{\displaystyle \frac{3}{2}}{t}^{2}dt}{t}=\frac{3}{4}\int (t^4-2t)dt=\frac{3}{4}\left(\frac{t^5}{5}-t^2\right)+C \\ =\frac{3}{4}\left(\frac{\sqrt[3]{{(2x+2)}^5}}{5}-\sqrt[3]{{(2x+2)}^2}\right)+C \end{gathered}$$

Chọn C

Ta có 

$$\begin{gathered} K=\int \frac{x(\sqrt{5x+3}-\sqrt{x+3})dx}{5x+3-x-3} \\ =\frac{1}{4}\int \left(\sqrt{5x+3}-\sqrt{x+3}\right)dx \\ =\frac{1}{6}\left(\frac{{\displaystyle 1}}{{\displaystyle 5}}\sqrt{{(5x+3)}^3}-\sqrt{{(x+3)}^3}\right)+C \end{gathered}$$

Cho C = -10, ta được một nguyên hàm là

 $\frac{1}{6}\left(\frac{{\displaystyle 1}}{{\displaystyle 5}}\sqrt{{(5x+3)}^3}-\sqrt{{(x+3)}^3}\right)-10$

Chọn A

Đặt $t=\cos x$$\Rightarrow dt=-\sin xdx$

Ta có

$$\begin{gathered} I=\int \left(1-{\cos }^{2}x\right){\cos }^{5}x\sin xdx=-\int (1-t^2)t^{5}dt \\ =\int (t^7-t^5)dt=\frac{t^8}{8}-\frac{t^6}{6}+C \\ =\frac{{\sin }^{8}x}{8}-\frac{{\sin }^{6}x}{6}+C \end{gathered}$$

Chọn B

$I=\int \frac{\cos xdx}{{\cos }^{3}x(\tan x+2{)}^3}=\int \frac{dx}{{\cos }^{2}x(\tan x+2{)}^3}$

Đặt $t=\tan x$$\Rightarrow dt=\frac{1}{{\cos }^{2}x}dx$

I = $\int \frac{dt}{{\left(t+2\right)}^3}$

Do đó $I=-\frac{1}{2}\frac{1}{{(\tan x+2)}^2}+C$

Chọn C

Đặt $u=x^2+1\Rightarrow du=2xdx$$\Rightarrow xdx=\frac{1}{2}du$

$$\begin{gathered} \Rightarrow \int \sqrt{x^2+1}.xdx=\int u^{\frac{1}{2}}.\frac{1}{2}du \\ =\frac{1}{2}\int u^{\frac{1}{2}}du=\frac{1}{2}{u}^{\frac{3}{2}}.\frac{2}{3}+C \\ =\frac{u^{{\displaystyle \frac{3}{2}}}}{3}+C=\frac{1}{3}\sqrt{{\left(x^2+1\right)}^3}+C \end{gathered}$$

Chọn A

Đặt $u=x^3+5\Rightarrow du=3x^{2}dx$$\Rightarrow x^{2}dx=\frac{1}{3}du$

$$\begin{gathered} \Rightarrow \int {\left(x^3+5\right)}^4{x}^{2}dx=\int \frac{1}{3}{u}^{4}du \\ =\frac{1}{3}\int u^{4}du=\frac{1}{3}.\frac{u^5}{5}+C \\ =\frac{u^5}{15}+C=\frac{{\left(x^3+5\right)}^5}{15}+C \end{gathered}$$

Chọn D

Đặt $u=x^2+5\Rightarrow du=2xdx$$\Rightarrow xdx=\frac{1}{2}du$

$$\begin{gathered} \Rightarrow \int \frac{x}{x^2+5}dx=\int \frac{1}{2}.\frac{1}{u}du \\ =\frac{1}{2}\ln \left|u\right|+C=\frac{1}{2}\ln \left(x^2+5\right)+C \end{gathered}$$

Chọn A

Đặt $u=x^2-2x+3\Rightarrow du=2(x-1)dx$$\Rightarrow (x-1)dx=\frac{du}{2}$

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

Chọn A

$\int \frac{\sin x}{\sqrt[3]{{\cos }^{2}x}}dx=\int \frac{\sin {\displaystyle x}}{{\displaystyle {\cos }^{\frac{2}{3}}}{\displaystyle x}}dx=\int \sin x.{\cos }^{-\frac{2}{3}}xdx$

Đặt $u=\cos x\Rightarrow du=-\sin xdx$

$$\begin{gathered} \Rightarrow \int \frac{\sin x}{\sqrt[3]{{\cos }^{2}x}}dx=-\int u^{-\frac{2}{3}}du \\ =-3u^{\frac{1}{3}}+C=-3\sqrt[3]{\cos x}+C \end{gathered}$$

Chọn B

Đặt $u=\mathrm{co}t2x\Rightarrow du=-\frac{2}{{\sin }^{2}2x}dx$$\Rightarrow du=-2\left(1+co{t}^{2}2x\right)dx$

$$\begin{gathered} \Rightarrow \int \left(1+co{t}^{2}2x\right)e^{cot2x}dx=-\frac{1}{2}\int e^{u}du \\ =-\frac{1}{2}{e}^{cot2x}+C \end{gathered}$$

Chọn A

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

Chọn C

$$\begin{gathered} \int f\left(x\right)dx=\int \sqrt[3]{3x+1}dx=\int {\left(3x+1\right)}^{\frac{1}{3}}\frac{d(3x+1)}{3} \\ =\frac{1}{3}\int {(3x+1)}^{\frac{1}{3}}d(3x+1)=\frac{1}{3}\frac{{(3x+1)}^{{\displaystyle \frac{4}{3}}}}{{\displaystyle \frac{4}{3}}}+C \\ \Rightarrow \int f\left(x\right)dx=\frac{1}{4}\left(3x+1\right)\sqrt[3]{3x+1}+C \end{gathered}$$

Chọn C

$$\begin{gathered} \int \left(\sqrt[3]{x^2}+\frac{14}{1-x}\right)dx=\int \left(x^{\frac{2}{3}}+14.\frac{1}{1-x}\right)dx \\ =\frac{3}{5}\sqrt[3]{x^5}-14\ln \left|1-x\right|+C \end{gathered}$$

Chọn D

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

Chọn A

$$\begin{gathered} \int \sin x.\cos 2xdx=\int \left(2{\cos }^{2}x-1\right)\sin xdx \\ =-\int \left(2{\cos }^{2}x-1\right)d\left(\cos x\right) \\ =\frac{-2{\cos }^{3}x}{3}+\cos x+C \end{gathered}$$

Chọn B

$$\begin{gathered} \int 2\sin x.\cos 3xdx=\int \left(\sin 4x-\sin 2x\right)dx \\ =\frac{1}{2}\cos 2x-\frac{1}{4}\cos 4x+C \end{gathered}$$