Bài tập nguyên hàm từng phần có đáp án chi tiết siêu hay.

Bài tập nguyên hàm từng phần có đáp án chi tiết siêu hay.

Dưới đây là một số bài tập trắc nghiệm nguyên hàm từng phần có Lời giải chi tiết

Lời giải chi tiết:

a) ${{I}_{1}}=\int{x\sin xdx}$

• Cách 1: Đặt $\left\{ \begin{array}  {} u=x \\  {} \sin xdx=dv \\ \end{array} \right.\overset{{}}{\longleftrightarrow}\left\{ \begin{array}  {} du=dx \\  {} v=-\cos x \\ \end{array} \right.$

$\xrightarrow{{}}{{I}_{1}}=\int{x\sin xdx=-x\cos x+\int{\cos xdx=-x\cos x+\sin x+C.}}$ 

• Cách 2: ${{I}_{1}}=\int{x\sin xdx}=-\int{xd\left( \cos x \right)=-\left[ x\cos x-\int{\cos xdx} \right]=-x\cos x+\sin x+C}$

b) ${{I}_{2}}=\int{x{{e}^{3x}}dx}$

• Cách 1: Đặt $\left\{ \begin{array}  {} u=x \\  {} {{e}^{3x}}dx=dv \\ \end{array} \right.\overset{{}}{\longleftrightarrow}\left\{ \begin{array}  {} du=dx \\  {} v=\frac{1}{3}{{e}^{3x}} \\ \end{array} \right.$

$\xrightarrow{{}}{{I}_{2}}=\int{x{{e}^{3x}}dx=\frac{1}{3}x{{e}^{3x}}-\frac{1}{3}\int{{{e}^{3x}}dx}=\frac{1}{3}x{{e}^{3x}}-\frac{1}{9}\int{{{e}^{3x}}d\left( 3x \right)=\frac{1}{3}x{{e}^{3x}}-\frac{1}{9}{{e}^{3x}}+C}}$ 

• Cách 2: 

${{I}_{2}}=\int{x{{e}^{3x}}dx}=\frac{1}{3}\int{xd\left( {{e}^{3x}} \right)=\frac{1}{3}\left[ x{{e}^{3x}}-\int{{{e}^{3x}}dx} \right]=\frac{1}{3}\left[ x{{e}^{3x}}-\frac{1}{3}\int{{{e}^{3x}}d\left( 3x \right)} \right]=\frac{1}{3}\left( x{{e}^{3x}}-\frac{1}{3}{{e}^{3x}} \right)}+C$

c) ${{I}_{3}}=\int{{{x}^{2}}\cos xdx}$

• Cách 1: Đặt $\left\{ \begin{array}  {} u={{x}^{2}} \\  {} \cos xdx=dv \\ \end{array} \right.\overset{{}}{\longleftrightarrow}\left\{ \begin{array}  {} du=2xdx \\  {} v=\sin x \\ \end{array} \right.$

Khi đó ${{I}_{3}}=\int{{{x}^{2}}\cos xdx={{x}^{2}}\sin x-\int{2x\sin xdx={{x}^{2}}\sin x-2J}}$

Xét $J=\int{x\sin xdx.}$ Đặt

$\left\{ \begin{array}  {} u=x \\  {} \sin xdx=dv\overset{{}}{\longleftrightarrow} \\ \end{array} \right.\left\{ \begin{array}  {} du=dx \\  {} v=-\cos x \\ \end{array} \right.\xrightarrow{{}}J=-x\cos x+\int{\cos xdx=-x\cos x+\sin x}$

$\xrightarrow{{}}{{I}_{3}}={{x}^{2}}\sin x-2\left( -x\cos x+\sin x \right)+C.$

• Cách 2: ${{I}_{3}}=\int{{{x}^{2}}\cos xdx=\int{{{x}^{2}}d\left( \sin x \right)={{x}^{2}}\sin x-\int{\sin xd\left( {{x}^{2}} \right)={{x}^{2}}\sin x-\int{2x\sin xdx}}}}$

$={{x}^{2}}\sin x+2\int{xd\left( \cos x \right)={{x}^{2}}\sin x}+2x\cos x-2\int{\cos xdx={{x}^{2}}\sin x+2x\cos x-2\sin x+C.}$ 

d) ${{I}_{4}}=\int{x\ln xdx}$

• Cách 1: Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} xdx=dv \\ \end{array} \right.\overset{{}}{\longleftrightarrow}\left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=\frac{{{x}^{2}}}{2} \\ \end{array} \right.\xrightarrow{{}}{{I}_{4}}=\int{x\ln xdx=\frac{{{x}^{2}}}{2}\ln x-\int{\frac{{{x}^{2}}}{2}.\frac{dx}{x}=\frac{{{x}^{2}}}{2}\ln x-\frac{{{x}^{2}}}{4}+C.}}$
• Cách 2: Ta có:

${{I}_{4}}=\int{x\ln xdx=\int{\ln xd\left( \frac{{{x}^{2}}}{2} \right)=\frac{{{x}^{2}}}{2}\ln x-\int{\frac{{{x}^{2}}}{2}d\left( \ln x \right)=\frac{{{x}^{2}}}{2}\ln x-\int{\frac{{{x}^{2}}}{2}\frac{dx}{x}=\frac{{{x}^{2}}}{2}}\ln x-\frac{{{x}^{2}}}{4}}+C.}}$

Lời giải chi tiết:

a) ${{I}_{5}}=\int{{{x}^{2}}\ln xdx}$

• Cách 1: 

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} {{x}^{2}}dx=dv \\ \end{array} \right.\overset{{}}{\longleftrightarrow}\left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=\frac{{{x}^{3}}}{3} \\ \end{array} \right.\xrightarrow{{}}{{I}_{5}}=\int{{{x}^{2}}\ln xdx=\frac{{{x}^{3}}}{3}\ln -\int{\frac{{{x}^{3}}}{3}.\frac{dx}{x}=\frac{{{x}^{3}}}{3}\ln x-\frac{{{x}^{3}}}{9}+C.}}$

• Cách 2: 

Ta có ${{I}_{5}}=\int{{{x}^{2}}\ln xdx=\int{\ln xd\left( \frac{{{x}^{3}}}{3} \right)=\frac{{{x}^{3}}}{3}\ln x-\int{\frac{{{x}^{3}}}{3}d\left( \ln x \right)=\frac{{{x}^{3}}}{3}\ln x}-\int{\frac{{{x}^{3}}}{3}\frac{dx}{x}}=\frac{{{x}^{3}}}{3}\ln x-\frac{{{x}^{3}}}{9}+C.}}$

b) ${{I}_{6}}=\int{x{{\ln }^{2}}\left( x+1 \right)dx}$

Ta có ${{I}_{6}}=\int{x{{\ln }^{2}}\left( x+1 \right)dx}=\int{{{\ln }^{2}}\left( x+1 \right)d\left( \frac{{{x}^{^{2}}}}{2} \right)=\frac{{{x}^{2}}}{2}{{\ln }^{2}}\left( x+1 \right)-\int{\frac{{{x}^{2}}}{2}d\left( {{\ln }^{2}}\left( x+1 \right) \right)}}$

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

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

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

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

Xét $K=\int{\frac{{{x}^{2}}-2x}{x+1}dx}=\int{\left( x-3+\frac{3}{x+1} \right)dx}=\frac{{{x}^{2}}}{2}-3x+3\ln \left| x+1 \right|$

$\xrightarrow{{}}J=\left( \frac{{{x}^{2}}}{2}-x \right)\ln \left( x+1 \right)-\frac{1}{2}\left( \frac{{{x}^{2}}}{2}-3x+3\ln \left| x+1 \right| \right)+\frac{{{\ln }^{2}}\left( x+1 \right)}{2}+C.$

Từ đó ta được ${{I}_{6}}=\frac{{{x}^{2}}{{\ln }^{2}}\left( x+1 \right)}{2}-\left( \frac{{{x}^{2}}}{2}-x \right)\ln \left( x+1 \right)+\frac{1}{2}\left( \frac{{{x}^{2}}}{2}-3x+3\ln \left| x+1 \right| \right)-\frac{{{\ln }^{2}}\left( x+1 \right)}{2}+C.$

c) ${{I}_{7}}=\int{\ln \left( x+\sqrt{1+{{x}^{2}}} \right)dx}$ 

Ngầm hiểu $u=\ln \left( x+\sqrt{1+{{x}^{2}}} \right);v=x$ ta có

${{I}_{7}}=x\ln \left( x+\sqrt{1+{{x}^{2}}} \right)-\int{xd\left[ \ln \left( x+\sqrt{1+{{x}^{2}}} \right) \right]=x\ln \left( x+\sqrt{1+{{x}^{2}}} \right)-\int{\frac{1+\frac{x}{\sqrt{1+{{x}^{2}}}}}{x+\sqrt{1+{{x}^{2}}}}xdx}}$

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

Vậy ${{I}_{7}}=x\ln \left( x+\sqrt{1+{{x}^{2}}} \right)-\sqrt{1+{{x}^{2}}}+C.$

d) ${{I}_{8}}=\int{{{e}^{x}}\sin xdx}$

${{I}_{8}}=\int{{{e}^{x}}\sin xdx=\int{\sin xd\left( {{e}^{x}} \right)={{e}^{x}}\sin x-\int{{{e}^{x}}d\left( \sin x \right)={{e}^{x}}\sin x-\int{{{e}^{x}}\cos xdx={{e}^{x}}\sin x-\int{\cos xd\left( {{e}^{x}} \right)}}}}}$

$={{e}^{x}}\sin x-\int{\cos xd\left( {{e}^{x}} \right)}={{e}^{x}}\sin x-\left[ {{e}^{x}}\cos x-\int{{{e}^{x}}d\left( \cos x \right)} \right]={{e}^{x}}\sin x-\left[ {{e}^{x}}\cos x+\int{{{e}^{x}}\sin xdx} \right]$

$={{e}^{x}}\sin x-\left[ {{e}^{x}}\cos x+{{I}_{8}} \right]={{e}^{x}}\sin x-{{e}^{x}}\cos x-{{I}_{8}}\xrightarrow{{}}{{I}_{8}}=\frac{{{e}^{x}}\sin x-{{e}^{x}}\cos x}{2}+C.$

Nhận xét: Trong nguyên hàm ${{I}_{8}}$ chúng ta thấy rất rõ là việc tính nguyên hàm gồm hai vòng lặp, trong mỗi vòng ta đều nhất quán đặt $u$ là hàm lượng giác (sinx hoặc cosx) và việc tính toán không thể tính trực tiếp được. 

Lời giải chi tiết

a) Đặt $\left\{ \begin{array}  {} u=\ln \left( x-1 \right) \\  {} dv=\frac{1}{{{\left( 2x+1 \right)}^{2}}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{1}{x-1}dx \\  {} v=\frac{-1}{2\left( 2x+1 \right)} \\ \end{array} \right..$ Khi đó: ${{I}_{9}}=\frac{-\ln \left( x-1 \right)}{2\left( 2x+1 \right)}+\int{\frac{dx}{2\left( 2x+1 \right)\left( x-1 \right)}}$

$\frac{-\ln \left( x-1 \right)}{2\left( 2x+1 \right)}+\frac{1}{6}\int{\left( \frac{1}{x-1}-\frac{2}{2x+1} \right)dx=\frac{-\ln \left( x-1 \right)}{2\left( 2x+1 \right)}+\frac{1}{6}\ln \left| \frac{x-1}{2x+1} \right|+C}$

b) Đặt $\left\{ \begin{array}  {} u=\ln \left( 2x+1 \right) \\  {} dv=\frac{1}{{{\left( 1-3x \right)}^{2}}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{1}{2x+1}dx \\  {} v=\frac{-1}{3\left( 3x-1 \right)} \\ \end{array} \right..$ Khi đó: ${{I}_{10}}=\frac{-\ln \left( 2x+1 \right)}{3\left( 3x-1 \right)}+\int{\frac{dx}{3\left( 2x+1 \right)\left( 3x-1 \right)}}$

$-\frac{\ln \left( 2x+1 \right)}{3\left( 3x-1 \right)}+\frac{1}{15}\int{\left( \frac{3}{3x-1}-\frac{2}{2x+1} \right)dx=\frac{-\ln \left( 2x+1 \right)}{3\left( 3x-1 \right)}+\frac{1}{15}\ln \left| \frac{3x-1}{2x+1} \right|+C}$

c) Đặt $\left\{ \begin{array}  {} u=x \\  {} dv=\sin x{{\cos }^{2}}xdx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=dx \\  {} v=\frac{-{{\cos }^{3}}x}{3} \\ \end{array} \right..$ Khi đó ${{I}_{11}}=\frac{-x{{\cos }^{3}}x}{3}+\frac{1}{3}\int{{{\cos }^{3}}xdx}$

$=\frac{-x{{\cos }^{3}}x}{3}+\frac{1}{3}\int{\frac{\cos 3x+3\cos x}{4}dx=\frac{-x{{\cos }^{3}}x}{3}+\frac{\sin 3x}{36}+\frac{\sin x}{4}+C}$

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

$\Rightarrow {{I}_{12}}=\frac{-{{x}^{2}}{{e}^{x}}}{x+2}+\int{x{{e}^{x}}dx}=\frac{-{{x}^{2}}{{e}^{x}}}{x+2}+\int{x{{e}^{x}}dx}=-\frac{{{x}^{2}}{{e}^{x}}}{x+2}+x{{e}^{x}}-{{e}^{x}}+C.$

Lời giải chi tiết

a) ${{I}_{13}}=\int{x\ln \left( {{x}^{2}}+1 \right)dx.}$ Đặt $\left\{ \begin{array}  {} u=\ln \left( 1+{{x}^{2}} \right) \\  {} xdx=dv \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{2xdx}{1+{{x}^{2}}} \\  {} v=\frac{{{x}^{2}}+1}{2} \\ \end{array} \right.$

$\Rightarrow {{I}_{13}}=\int{x\ln \left( 1+{{x}^{2}} \right)dx}=\frac{\left( {{x}^{2}}+1 \right)}{2}\ln \left( 1+{{x}^{2}} \right)-\int{\frac{\left( {{x}^{2}}+1 \right)}{2}}\frac{2x}{1+{{x}^{2}}}dx=\frac{\left( {{x}^{2}}+1 \right)}{2}\ln \left( 1+{{x}^{2}} \right)-\int{xdx}$

$\Rightarrow {{I}_{13}}=\frac{\left( {{x}^{2}}+1 \right)}{2}\ln \left( 1+{{x}^{2}} \right)-\frac{{{x}^{2}}}{2}+C=\frac{\left( {{x}^{2}}+1 \right)\ln \left( 1+{{x}^{2}} \right)-{{x}^{2}}}{2}+C$

b) ${{I}_{14}}=\int{x{{\tan }^{2}}xdx=\int{x\left( 1-\frac{1}{{{\cos }^{2}}x} \right)dx=\int{xdx}-\int{\frac{x}{{{\cos }^{2}}x}}dx}}$

Ta đi tính $J=\int{\frac{x}{{{\cos }^{2}}x}dx.}$ Đặt $\left\{ \begin{array}  {} u=x \\  {} \frac{1}{{{\cos }^{2}}x}dx=dv \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=dx \\  {} v=\tan x \\ \end{array} \right.$

$\Rightarrow J=x\tan x-\int{\tan xdx=x\tan x-\int{\frac{\sin xdx}{\cos x}=x\tan x+\int{\frac{d\left( \cos x \right)}{\cos x}}=x\tan x+\ln \left| \cos x \right|+C}}$

$\Rightarrow {{I}_{14}}=\frac{{{x}^{2}}}{2}+x\tan x+\ln \left| \cos x \right|+C$

c) ${{I}_{15}}=\int{{{x}^{2}}\ln \left( {{x}^{2}}+1 \right)dx.}$ Đặt $\left\{ \begin{array}  {} u=\ln \left( 1+{{x}^{2}} \right) \\  {} {{x}^{2}}dx=dv \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{2xdx}{1+{{x}^{2}}} \\  {} v=\frac{{{x}^{3}}}{3} \\ \end{array} \right.$

$\Rightarrow {{I}_{15}}=\int{{{x}^{2}}\ln \left( {{x}^{2}}+1 \right)dx=\frac{{{x}^{3}}}{3}\ln \left( {{x}^{2}}+1 \right)-\int{\frac{{{x}^{3}}}{3}.\frac{2x}{{{x}^{2}}+1}dx}=\frac{{{x}^{3}}}{3}\ln \left( {{x}^{2}}+1 \right)-\frac{2}{3}\int{\frac{{{x}^{4}}}{{{x}^{2}}+1}dx}}$

Ta đi tính $K=\int{\frac{{{x}^{4}}}{1+{{x}^{2}}}}dx$

Đặt $x=\tan t\Rightarrow dx=\frac{dt}{{{\cos }^{2}}t}$ và ${{x}^{2}}+1={{\tan }^{2}}x+1=\frac{1}{{{\cos }^{2}}t}\Rightarrow K=\int{\frac{{{x}^{4}}}{1+{{x}^{2}}}dx=\arctan x+\frac{{{x}^{3}}-3x}{3}+C}$

Do đó: ${{I}_{15}}=\int{{{x}^{2}}\ln \left( {{x}^{2}}+1 \right)dx=\frac{{{x}^{3}}\ln \left( {{x}^{2}}+1 \right)}{3}+\frac{2}{3}\left( \arctan x+\frac{{{x}^{3}}-3x}{3} \right)+C}$

d) ${{I}_{16}}=\int{\sqrt{x}\sin \sqrt{x}dx}$

Đặt $\sqrt{x}=t\Rightarrow dt=\frac{1}{2}\sqrt{x}dx\Rightarrow 2dt=dx\Rightarrow {{I}_{16}}=\int{2t\sin tdt.}$ Đặt $\left\{ \begin{array}  {} u=t \\  {} \sin tdt=dv \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=dt \\  {} v=-\cos t \\ \end{array} \right.$

$\Rightarrow {{I}_{16}}=\int{2t\sin tdt=2\left[ -t\cos t+\int{\cos t} \right]=-2t\cos t+2\sin t+C\Rightarrow {{I}_{16}}=-2\sqrt{x}\cos \sqrt{x}}+2\sin \sqrt{x}+C$

Lời giải chi tiết:

Đặt $\left\{ \begin{array}  {} u=\ln \left( x+2 \right) \\  {} dv=dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{dx}{x+2} \\  {} v=x+2 \\ \end{array} \right.$ (Ta có thể chọn $v=x;v=x+1...,$ tuy nhiên ta nên chọn $v=x+2$ để tính toán dễ dàng hơn).

Khi đó $I=\left( x+2 \right)\ln \left( x+2 \right)-\int{dx}=\left( x+2 \right)\ln \left( x+2 \right)-x+C.$ Chọn B.

Lời giải chi tiết:

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

Khi đó $I=\frac{{{x}^{2}}-1}{2}\ln \left( x-1 \right)-\int{\frac{x+1}{2}dx=\frac{{{x}^{2}}-1}{2}\ln \left( x-1 \right)-\frac{{{x}^{2}}}{4}-\frac{x}{2}+C.}$ Chọn D.

Lời giải chi tiết:

Đặt $\left\{ \begin{array}  {} u=x-2 \\  {} dv={{e}^{x}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=dx \\  {} v={{e}^{x}} \\ \end{array} \right.\Rightarrow I=\left( x-2 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x-2 \right){{e}^{x}}-{{e}^{x}}+C=\left( x-3 \right){{e}^{x}}+C.$ Chọn A.

Lời giải chi tiết:

Ta có: $F\left( x \right)=\int{\left( 2x+1 \right)\sin xdx.}$ Đặt $\left\{ \begin{array}  {} u=2x+1 \\  {} dv=\sin xdx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=2dx \\  {} v=-\cos x \\ \end{array} \right.$

$\Rightarrow F\left( x \right)=-\left( 2x+1 \right)\cos x+\int{2\sin xdx=-\left( 2x+1 \right)\cos x+2\sin x+C}$

Mặt khác $F\left( 0 \right)=-1+C=3\Rightarrow C=4\Rightarrow F\left( x \right)=-\left( 2x+1 \right)\cos x+2\sin x+4.$ Chọn B.

Lời giải chi tiết:

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=\frac{dx}{{{\left( x+1 \right)}^{2}}} \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=-\frac{1}{x+1}+1=\frac{x}{x+1} \\ \end{array} \right.\Rightarrow I=\frac{x\ln x}{x+1}-\int{\frac{dx}{x+1}=\frac{x\ln x}{x+1}-\ln \left| x+1 \right|+C.}$ Chọn C.

Lời giải chi tiết:

Đặt $\left\{ \begin{array}  {} u=2-x \\  {} dv=\cos xdx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=-dx \\  {} v=\sin x \\ \end{array} \right.\Rightarrow I=\left( 2-x \right)\sin x+\int{\sin xdx=\left( 2-x \right)\sin x-\cos x+C.}$ Chọn B.

Lời giải chi tiết:

Đặt $\left\{ \begin{array}  {} u=x+1 \\  {} dv={{3}^{x}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=dx \\  {} v=\frac{{{3}^{x}}}{\ln 3} \\ \end{array} \right.\Rightarrow I=\frac{\left( x+1 \right){{3}^{x}}}{\ln 3}-\int{\frac{{{3}^{x}}dx}{\ln 3}}\Rightarrow I=\frac{\left( x+1 \right){{3}^{x}}}{\ln 3}-\frac{{{3}^{x}}}{{{\ln }^{2}}3}+C.$ Chọn D.

Lời giải chi tiết:

Ta có: $I=\int{x}\frac{1+\cos 2x}{2}dx=\frac{1}{2}\int{xdx}+\frac{1}{2}\int{x\cos 2xdx}$

Đặt $\left\{ \begin{array}  {} u=x \\  {} dv=\cos 2xdx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=dx \\  {} v=\frac{\sin 2x}{2} \\ \end{array} \right.\Rightarrow \int{x\cos 2xdx=\frac{x\sin 2x}{2}-\int{\frac{\sin 2xdx}{2}=\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+C}}$

$\Rightarrow I=\frac{1}{4}{{x}^{2}}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+C\Rightarrow m+n+p=\frac{5}{8}.$ Chọn D.

Lời giải:

Tính nguyên hàm $I=\int{f'\left( x \right)\tan xdx}$

Đặt $\left\{ \begin{array}  {} u=\tan x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{dx}{{{\cos }^{2}}x} \\  {} v=f\left( x \right) \\ \end{array} \right.\Rightarrow I=f\left( x \right).\tan x-\int{\frac{f\left( x \right)dx}{{{\cos }^{2}}x}=f\left( x \right)\tan x-\frac{1}{{{x}^{2}}}+C}$

Mặt khác $\frac{f\left( x \right)}{{{\cos }^{2}}x}=F'\left( x \right)=\frac{-2x}{{{x}^{4}}}=\frac{-2}{{{x}^{3}}}\Rightarrow f\left( x \right)=\frac{-2{{\cos }^{2}}x}{{{x}^{3}}}$

Do đó $I=\frac{-2{{\cos }^{2}}x}{{{x}^{3}}}.\tan x-\frac{1}{{{x}^{2}}}+C=\frac{-\sin 2x}{{{x}^{3}}}-\frac{1}{{{x}^{2}}}+C.$ Chọn A.

Lời giải:

Tính nguyên hàm $I=\int{f'\left( x \right)\cos xdx}$

Đặt $\left\{ \begin{array}  {} u=\cos x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=-\sin xdx \\  {} v=f\left( x \right) \\ \end{array} \right.$

$\Rightarrow I=f\left( x \right).\cos x+\int{f\left( x \right)\sin xdx=f\left( x \right)\cos x+\left( 1-\frac{{{x}^{2}}}{2} \right)\cos x+}x\sin x$

Mặt khác $F'\left( x \right)=-x\cos x-\left( 1-\frac{{{x}^{2}}}{2} \right)\sin x+\sin x+x\cos x=\frac{{{x}^{2}}\sin x}{2}=f\left( x \right)\sin x$

Do đó $f\left( x \right)=\frac{{{x}^{2}}}{2}\Rightarrow I=\cos x+x\sin x.$ Chọn C.

Lời giải:

Tính nguyên hàm $I=\int{f'\left( x \right)\ln xdx}$

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=f\left( x \right) \\ \end{array} \right.\Rightarrow I=f\left( x \right)\ln x-\int{\frac{f\left( x \right)dx}{x}=f\left( x \right)\ln x-{{e}^{x}}-x+C.}$

Mặt khác $\frac{f\left( x \right)}{x}=F'\left( x \right)={{e}^{x}}+1\Rightarrow f\left( x \right)=x\left( {{e}^{x}}+1 \right)$

Suy ra $I=x\left( {{e}^{x}}+1 \right)\ln x-{{e}^{x}}-x+C.$ Chọn B.

Lời giải:

Đặt $\left\{ \begin{array}  {} u={{e}^{x}} \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du={{e}^{x}}dx \\  {} v=f\left( x \right) \\ \end{array} \right.\Rightarrow I=\int{f'\left( x \right){{e}^{x}}dx={{e}^{x}}.f\left( x \right)-\int{f\left( x \right).{{e}^{x}}dx}}$

$=f\left( x \right){{e}^{x}}-x\sin x+C.$

Lại có: $f\left( x \right).{{e}^{x}}=F'\left( x \right)=\sin x+x\cos x$

$\Rightarrow I=\sin x+x\cos x-x\sin x+C=x\left( \cos x-\sin x \right)+\sin x+C.$ Chọn D.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=f\left( x \right) \\ \end{array} \right.$ suy ra $\int{f'\left( x \right).\ln xdx=\ln x.f\left( x \right)-\int{\frac{f\left( x \right)}{x}dx}}$

Ta có ${F}'\left( x \right)=\frac{f\left( x \right)}{x}\Leftrightarrow 2x=\frac{f\left( x \right)}{x}\Leftrightarrow f\left( x \right)=2{{x}^{2}}$

Do đó $\int{{f}'\left( x \right).\ln xdx=2{{x}^{2}}.\ln x-{{x}^{2}}-1+C={{x}^{2}}\left( 2\ln x-1 \right)+C.}$ Chọn D.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=f\left( x \right) \\ \end{array} \right.$ suy ra $\int{f'\left( x \right).\ln xdx=\ln x.f\left( x \right)-\int{\frac{f\left( x \right)}{x}dx}}$

Ta có ${F}'\left( x \right)=x.f\left( x \right)\Leftrightarrow \frac{1}{x}=x.f\left( x \right)\Leftrightarrow f\left( x \right)=\frac{1}{{{x}^{2}}}$

Do đó $\int{{f}'\left( x \right).\ln xdx=\frac{\ln x}{{{x}^{2}}}-\int{\frac{dx}{{{x}^{3}}}+C=\frac{\ln x}{{{x}^{2}}}+\frac{1}{2{{x}^{2}}}+C}.}$ Chọn A.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=f\left( x \right) \\ \end{array} \right.$ suy ra $\int{f'\left( x \right).\ln xdx=\ln x.f\left( x \right)-\int{\frac{f\left( x \right)}{x}dx}}$

Ta có ${F}'\left( x \right)=\frac{f\left( x \right)}{{{x}^{3}}}\Leftrightarrow \frac{1}{x}=\frac{f\left( x \right)}{{{x}^{3}}}\Leftrightarrow f\left( x \right)={{x}^{2}}$

Do đó $\int{{f}'\left( x \right).\ln xdx={{x}^{2}}\ln x-\int{xdx={{x}^{2}}.\ln x-\frac{{{x}^{2}}}{2}+C}.}$ Chọn D.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=\tan x \\  {} dv=f'\left( x \right)dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=\frac{dx}{{{\cos }^{2}}x} \\  {} v=f\left( x \right) \\ \end{array} \right.\Rightarrow \int{f'\left( x \right).tanxdx=f\left( x \right).\tan x-\int{\frac{f\left( x \right)}{{{\cos }^{2}}x}dx}}$

Ta có ${F}'\left( x \right)=\frac{f\left( x \right)}{{{\cos }^{2}}x}\Leftrightarrow \cot x+\frac{x}{{{\cos }^{2}}x}-\tan x=\frac{f\left( x \right)}{{{\cos }^{2}}x}\Leftrightarrow f\left( x \right)=x.$

Do đó $\int{{f}'\left( x \right).tanxdx=x.\tan x-x.\tan x-\ln \left| \cos x \right|+C=-\ln \left| \cos x \right|+C.}$ Chọn C.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=\ln x \\  {} dv=dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=\frac{dx}{x} \\  {} v=x \\ \end{array} \right.$ suy ra $\int{f\left( x \right)dx=x.\ln x-\int{dx}=x.\ln x-x+C}$

Mà $F\left( 1 \right)=3\xrightarrow{{}}1.\ln 1-1+C=3\Leftrightarrow C=4.$ Vậy $T=17.$ Chọn D.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=x \\  {} dv={{e}^{2x}}dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=dx \\  {} v=\frac{{{e}^{2x}}}{2} \\ \end{array} \right.\Rightarrow \int{f\left( x \right)dx=\frac{x.{{e}^{2x}}}{2}-\int{\frac{{{e}^{2x}}}{2}dx}=\frac{x.{{e}^{2x}}}{2}-\frac{{{e}^{2x}}}{4}+C}$

Mà $F\left( \frac{1}{2} \right)=0\xrightarrow{{}}C=0\xrightarrow{{}}F\left( x \right)=\frac{x.{{e}^{2x}}}{2}-\frac{{{e}^{2x}}}{4}.$ Vậy $\ln \left| F\left( \frac{5}{2} \right) \right|=5.$ Chọn C.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=x \\  {} dv={{e}^{-x}}dx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=dx \\  {} v=-{{e}^{-x}} \\ \end{array} \right.\Rightarrow \int{f\left( x \right)dx=-x.{{e}^{-x}}+\int{{{e}^{-x}}dx}=-x.{{e}^{-x}}-{{e}^{-x}}+C}$

Mà $F\left( 0 \right)=-1\xrightarrow{{}}C-1=-1\Leftrightarrow C=0\xrightarrow{{}}F\left( x \right)=-x.{{e}^{-x}}-{{e}^{-x}}.$

Do đó $F\left( x \right)+x+1=0\Leftrightarrow -x.{{e}^{-x}}-{{e}^{-x}}+x+1=0\Leftrightarrow \left( x+1 \right)\left( 1-{{e}^{-x}} \right)=0\Leftrightarrow \left[ \begin{array}  {} x=-1 \\  {} x=0 \\ \end{array} \right..$ Chọn D.

Lời giải:

Đặt $\left\{ \begin{array}  {} u=x \\  {} dv=\sin xdx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du=dx \\  {} v=-\cos x \\ \end{array} \right.\Rightarrow \int{x.\sin xdx=-x.\cos x+\int{\cos xdx}=-x.\cos x+\sin x+C}$

Mà $F\left( \pi  \right)=2\pi \xrightarrow{{}}C=4\pi .$ Do đó $F\left( x \right)=-x.\cos x+\sin x+4\pi .$

Vậy $T=2.4\pi -8.2\pi =-8\pi .$ Chọn C.