150 câu trắc nghiệm Nguyên hàm - Tích phân cơ bản (P4)

Chọn C

Đặt $t=x^4-1\Rightarrow dt=4x^{3}dx$

Đổi cận: khi x = 0 => t = -1; x = 1 => t = 0

$$\begin{gathered} B={\int }_{-1}^0\frac{1}{4}{t}^{5}dt=\frac{1}{4}.\frac{t^6}{6}{|}_{-1}^0 \\ =\frac{1}{24}{t}^6{|}_{-1}^0=\frac{-1}{24} \end{gathered}$$

Chọn D

Đặt $t=e^x-1\Rightarrow dt=e^{x}dx$

Đổi cận: khi x = 1 => t = e-1; x = 2 => t = $e^2-1$

$$\begin{gathered} C={\int }_{e-1}^{e^2-1}\frac{dt}{t}=\ln \left|t\right|{|}_{e-1}^{e^2-1} \\ =\ln (e^2-1)-\ln (e-1) \\ =\ln \frac{e^2-1}{e-1}=\ln (e+1) \end{gathered}$$

Chọn D

Đặt $t=4-x^2\Rightarrow dt=-2xdx$$\Rightarrow xdx=-\frac{dt}{2}$

Khi x = 0 => t = 4; x = 2 => t = 0

$$\begin{gathered} D={\int }_4^0\left(-\frac{1}{2}\sqrt{t}\right)dt=\frac{1}{2}{\int }_0^4{t}^{\frac{1}{2}}dt \\ =\frac{1}{2}\left(\frac{2}{3}{t}^{\frac{3}{2}}\right)\overline{)\begin{array}{c}4\\ 0\end{array}=\frac{1}{3}\left(t\sqrt{t}\right)}\overline{)\begin{array}{c}4\\ 0\end{array}} \\ =\frac{1}{3}(4.2-0)=\frac{8}{3} \end{gathered}$$

Chọn B

Áp dụng phương pháp đồng nhất hệ số ta có:

$\frac{-x+8}{x^2+5x+4}=\frac{-x+8}{(x+1)(x+4)}=\frac{3}{x+1}-\frac{4}{x+4}$

$$\begin{gathered} {\int }_0^3\frac{-x+8}{x^2+5x+4}dx={\int }_0^3\left(\frac{3}{x+1}-\frac{4}{x+4}\right)dx \\ =\left(3\ln \left(x+1\right)-4\ln \left(x+4\right)\right){|}_0^3=7\ln 4-4\ln 7 \end{gathered}$$

$\Rightarrow \left\{\begin{array}{l}a=7\\ b=4\end{array}\Rightarrow {\left(\frac{b}{a}\right)}^2=\frac{16}{49}\right.$

Chọn B

$$\begin{gathered} {\int }_1^2\frac{dx}{4x^2-4x+1}={\int }_1^2\frac{{\displaystyle dx}}{{(2x-1)}^2} \\ =-\frac{1}{2}.\frac{1}{2x-1}\overline{)\begin{array}{c}2\\ 1\end{array}=-\frac{1}{2}.\left(\frac{1}{3}-1\right)=\frac{1}{2}-\frac{1}{6}} \end{gathered}$$

$\Rightarrow \left\{\begin{array}{l}a=2\\ b=6\end{array}\right.\Rightarrow a,b$ là nghiệm của phương trình $x^2-8x+12=0$

Chọn D

Áp dụng phương pháp đồng nhất hệ số ta có:

$\frac{x}{4-x^2}=\frac{-x}{(x-2)(x+2)}=-\frac{1}{2}\left(\frac{1}{x-2}+\frac{1}{x+2}\right)$

$$\begin{gathered} I={\int }_0^1\frac{x}{4-x^2}dx=-\frac{1}{2}{\int }_0^1\left(\frac{1}{x-2}+\frac{1}{x+2}\right)dx \\ =-\frac{1}{2}\ln \left|\frac{x-2}{x+2}\right|\overline{)\begin{array}{c}1\\ 0\end{array}=-\frac{1}{2}\ln \frac{1}{3}} \\ \Rightarrow \left\{\begin{array}{l}a=1\\ b=3\end{array}\Rightarrow a^2-b=-2\right. \end{gathered}$$

Chọn D

Cho $x^2-3x+2=0\iff [\begin{array}{c}x=1\\ x=2\end{array}$(thỏa mãn)

Bảng xét dấu của $x^2-3x+2$ trên đoạn $\left[0;2\right]\begin{array}{}\end{array}$

Khi đó: 

$$\begin{gathered} I={\int }_0^1(x^2-3x+2)dx-{\int }_1^2(x^2-3x+2)dx \\ =\left(\frac{x^3}{3}-\frac{3x^2}{2}+2x\right)\overline{)\begin{array}{c}1\\ 0\end{array}-\left(\frac{x^3}{3}-\frac{3x^2}{2}+2x\right)\overline{)\begin{array}{c}2\\ 1\end{array}}} \\ =\frac{5}{6}-\left(-\frac{1}{6}\right)=1 \end{gathered}$$

Chọn A

Đặt $t=\sqrt{x}\Rightarrow dt=\frac{1}{2\sqrt{x}}dx$$\Rightarrow \frac{dx}{\sqrt{x}}=2dt$

Khi x=1 => t=1; x=4 => t=2

$\Rightarrow E={\int }_1^{2}2.e^{t}dt=2e^t{|}_1^2=2(e^2-e)$

Chọn B

Đặt $t={\sin }^{2}x\Rightarrow dt=2\sin x\cos xdx=\sin 2xdx$

Khi $x=0\Rightarrow {\sin }^{2}0=0\Rightarrow t=0$; $x=\frac{\mathrm{\pi }}{2}\Rightarrow {\sin }^2\frac{\mathrm{\pi }}{2}=1\Rightarrow t=1$

$\Rightarrow F={\int }_0^1\frac{dt}{1+t}=\ln \left|1+t\right|{|}_0^1=\ln 2-\ln 1=\ln 2$

Chọn C

Đặt $t=e^x-1\Rightarrow dt=e^{x}dx$

Đổi cận: Khi x=0 => t=0; x=ln2 => t=1

$\Rightarrow G={\int }_0^1{t}^{2}dt=\frac{t^3}{3}\overline{)\begin{array}{c}1\\ 0\end{array}=\frac{1}{3}}$

Chọn C

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

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

Suy ra $\left\{\begin{array}{l}a=\frac{1}{2}\\ b=-\frac{1}{4}\end{array}\right.\Rightarrow a.b=-\frac{1}{8}$

Chọn C

$$\begin{gathered} {\int }_0^1\frac{x+2}{x^2+4x+7}dx=\frac{1}{2}{\int }_0^1\frac{d(x^2+4x+7)}{x^2+4x+7} \\ =\frac{1}{2}\ln \left|x^2+4x+7\right|{|}_0^1 \\ =\frac{1}{2}\ln 12-\frac{1}{2}\ln 7=\ln \sqrt{12}-\ln \sqrt{7} \\ =a\ln \sqrt{12}+b\ln \sqrt{7} \end{gathered}$$

Do $a,b\in \mathrm{\mathbb{Z}}$$\Rightarrow \left\{\begin{array}{l}a=1\\ b=-1\end{array}\Rightarrow a+b=0\right.$

Chọn D

Vì $x^4+x^2+1>0,\forall x\in \left[-1;2018\right]$

$$\begin{gathered} \Rightarrow {\int }_{-1}^{2018}\left|x^4+x^2+1\right|dx \\ ={\int }_{-1}^{2018}(x^4+x^2+1)dx \end{gathered}$$

Chọn C

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\left(2\cos x-\sin 2x\right)dx \\ =2{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2xdx \\ =2\sin x\overline{)\begin{array}{c}\frac{\mathrm{\pi }}{2}\\ 0\end{array}+\frac{1}{2}\cos 2x\overline{)\begin{array}{c}\frac{\mathrm{\pi }}{2}\\ 0\end{array}=1}} \end{gathered}$$

Chọn C

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 3x \cos xdx \\ =\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\sin 4x+\sin 2x\right]dx \\ =\frac{1}{2}\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 4xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2xdx\right] \\ =\frac{1}{2}\left[-\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x\right]\overline{)\begin{array}{c}\frac{\mathrm{\pi }}{2}\\ 0\end{array}} \\ =\frac{1}{2}\left[\left(-\frac{1}{4}\cos 2\mathrm{\pi }-\frac{1}{2}\mathrm{cos\pi }\right)-\left(-\frac{1}{4}\cos 0-\frac{1}{2}\cos 0\right)\right] \\ =\frac{1}{2}\left(-\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}\right)=\frac{1}{2} \end{gathered}$$

Chọn C

$I={\int }_{\sqrt{3}}^{2\sqrt{2}}\frac{\sqrt{1+x^2}}{x}dx$

Đặt $t=\sqrt{1+x^2}$$\Rightarrow x^2=t^2-1$$\Rightarrow xdx=tdt$

Đổi cận: $x=\sqrt{3}\Rightarrow t=2$; $x=2\sqrt{2}\Rightarrow t=3$

$$\begin{gathered} I={\int }_2^3\frac{t.t}{t^2-1}dt={\int }_2^3(1+\frac{1}{(t-1)(t+1)})dt \\ =\left(t+\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|\right)\overline{)\begin{array}{c}3\\ 2\end{array}} \\ =3+\frac{1}{2}\ln \frac{1}{2}-2-\frac{1}{2}\ln \frac{1}{3}=1+\frac{1}{2}\ln \frac{3}{2}=1 +0,5(\ln 3-\ln 2) \end{gathered}$$

Chọn D

$J={\int }_{\sqrt{5}}^{2\sqrt{3}}\frac{xdx}{x^2\sqrt{x^2+4}}$

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

Đổi cận: $x=\sqrt{5}\Rightarrow t=3$; $x=2\sqrt{3}\Rightarrow t=4$

$$\begin{gathered} J={\int }_3^4\frac{tdt}{\left(t^2-4\right)t}={\int }_3^4\frac{dt}{t^2-4} \\ =\frac{1}{4}\ln \left|\frac{t-2}{t+2}\right| \overline{)\begin{array}{c}4\\ 3\end{array}=\frac{1}{4}\ln \frac{5}{3}} \end{gathered}$$

Chọn B

$I={\int }_{\ln 2}^{\ln 5}\frac{e^x.e^{x}dx}{\sqrt{e^x-1}}$

Đặt $t=\sqrt{e^x-1}$$\Rightarrow e^x=t^2+1$$\Rightarrow e^{x}dx=2t.dt$

Đổi cận: $x=\ln 2\Rightarrow t=1$; $x=\ln 5\Rightarrow t=2$

$$\begin{gathered} I=2{\int }_1^2\frac{\left(t^2+1\right)tdt}{t}=2{\int }_1^2\left(t^2+1\right)dt \\ =2\left(\frac{t^3}{3}+t\right) \overline{)\begin{array}{c}2\\ 1\end{array}=\frac{20}{3}} \end{gathered}$$

Chọn C

Ta có: $I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\left(1-{\cos }^{2}x\right)}^2\sin xdx$

Đặt $t=cosx\Rightarrow dt=-\sin xdx$

Đổi cận: $x=0\Rightarrow t=1$; $x=\frac{\mathrm{\pi }}{2}\Rightarrow t=0$

$$\begin{gathered} I=-{\int }_1^0{\left(1-t^2\right)}^{2}dt \\ ={\int }_0^1{\left(1-t^2\right)}^{2}dt \\ ={\int }_0^1\left(1-2t^2+t^4\right)dt=\frac{8}{15} \end{gathered}$$

Chọn C

Đặt $x=\sin t$ ta có $dx=costdt$.

Đổi cận: $x=0\Rightarrow t=0$; $x=1\Rightarrow t=\frac{\mathrm{\pi }}{2}$

Vậy

 $$\begin{gathered} I={\int }_0^1\sqrt{1-x^2}dx={\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\cos t\right|dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\cos tdt=\sin t {|}_0^{\frac{\mathrm{\pi }}{2}}=1 \end{gathered}$$

Chọn B

Đặt $x=\tan t$, ta có $dx=\left(1+{\tan }^{2}t\right)dt$

Đổi cận: $\left\{\begin{array}{l}x=0\to t=0\\ x=1\to t=\frac{\mathrm{\pi }}{4}\end{array}\right.$

Vậy $I={\int }_0^1\frac{dx}{1+x^2}={\int }_0^{\frac{\mathrm{\pi }}{4}}dt=t{|}_0^{\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{\pi }}{4}$

Chọn B

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

Vậy ${\int }_0^{1}x.e^{x}dx=x.e^x{|}_0^1-{\int }_0^1{e}^{x}dx$$=e-e^x{|}_0^1=e-\left(e-1\right)=1$

Chọn C

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

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{xdx}{{\cos }^{2}x}=\left(x \tan x\right) {|}_0^{\frac{\mathrm{\pi }}{4}}-{\int }_0^{\frac{\mathrm{\pi }}{4}}\tan xdx \\ =\frac{\mathrm{\pi }}{4}-{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin x}{\cos x}dx=\frac{\mathrm{\pi }}{4}+\left(\ln \left|\cos x\right|\right){|}_0^{\frac{\mathrm{\pi }}{4}} \\ =\frac{\mathrm{\pi }}{4}+\left(\ln \frac{\sqrt{2}}{2}-\ln 1\right)=\frac{\mathrm{\pi }}{4}+\ln \frac{\sqrt{2}}{2} \end{gathered}$$