Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x + 1}\\{dv = f\prime (x)dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = f(x)}\end{array}} \right.\)

\[ \Rightarrow F\left( x \right) = \left( {x + 1} \right)f\left( x \right) - \smallint f\left( x \right)dx + C\]

\[ \Rightarrow I = \smallint f\left( x \right)dx = \left( {x + 1} \right)f\left( x \right) - F\left( x \right) + C.\]

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = ln3x}\\{dv = {x^2}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{3}{{3x}}dx}\\{v = \frac{1}{3}{x^3}}\end{array}} \right.\)

\[ \Rightarrow I = \frac{1}{3}{x^3}\ln 3x - \smallint \frac{1}{3}{x^3}.\frac{3}{{3x}}dx = \frac{1}{3}{x^3}\ln 3x - \smallint \frac{1}{3}{x^2}dx = \frac{1}{3}{x^3}\ln 3x - \frac{1}{9}{x^3} + C\]

\[F\left( x \right) = \smallint \frac{x}{{{e^x}}}dx = \smallint x{e^{ - x}}dx\]

Đặt

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = {e^{ - x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - {e^{ - x}}}\end{array}} \right. \Rightarrow F(x) = - x{e^{ - x}} + \smallint {e^{ - x}}dx + C\\ = - x{e^{ - x}} - e - x + C = - (x + 1){e^{ - x}} + C = - \frac{{x + 1}}{{{e^x}}} + C.\end{array}\)

\[ - \frac{{x + a}}{{{e^x}}}\] là một họ nguyên hàm của hàm số

 \[f(x) = \frac{x}{{{e^x}}} \Rightarrow \left\{ {\begin{array}{*{20}{c}}{a = 1}\\{C = 0}\end{array}} \right.\]

\[F\left( x \right) = \smallint \frac{{{2^x} - 1}}{{{e^x}}}dx = \smallint \left( {{2^x} - 1} \right){e^{ - x}}dx = \smallint {2^x}{e^{ - x}}dx - \smallint {e^{ - x}}dx\]

\[ = \smallint {2^x}{e^{ - x}}dx + {e^{ - x}} + {C_1} = I + {e^{ - x}} + {C_1}.\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = {2^x}}\\{dv = {e^{ - x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = {2^x}ln2dx}\\{v = - {e^{ - x}}}\end{array}} \right.\)

\[\begin{array}{l} \Rightarrow I = - {2^x}{e^{ - x}} + ln2\smallint {2^x}{e^{ - x}}dx + {C_2} = - {2^x}{e^{ - x}} + ln2.I + {C_2}\\ \Leftrightarrow (ln2 - 1)I + {C_2} = {2^x}{e^{ - x}} \Rightarrow I = \frac{{{2^x}{e^{ - x}}}}{{ln2 - 1}} + {C_2}.\end{array}\]

\[ \Rightarrow F(x) = \frac{{{2^x}{e^{ - x}}}}{{ln2 - 1}} + {e^{ - x}} + C = \frac{{{2^x}}}{{(ln2 - 1){e^x}}} + \frac{1}{{{e^x}}} + C\]

\[ \Rightarrow F(0) = \frac{1}{{ln2 - 1}} + 1 + C = 1 \Rightarrow C = - \frac{1}{{ln2 - 1}}\]

\[ \Rightarrow F(x) = \frac{{{2^x}}}{{(ln2 - 1){e^x}}} + \frac{1}{{{e^x}}} - \frac{1}{{ln2 - 1}}\]

\[ = \frac{1}{{ln2 - 1}}{\left( {\frac{2}{e}} \right)^x} + {\left( {\frac{1}{e}} \right)^x} - \frac{1}{{ln2 - 1}}\]

\[I = \smallint x\sin x\cos xdx = \frac{1}{2}\smallint x\sin 2xdx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = sin2xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - \frac{{cos2x}}{2}}\end{array}} \right.\)

\[ \Rightarrow I = \frac{1}{2}\left( { - x.\frac{{\cos 2x}}{2} + \frac{1}{2}\smallint \cos 2xdx} \right) + C\]

\[ = \frac{1}{2}\left( { - \frac{{x\cos 2x}}{2} + \frac{{\sin 2x}}{4}} \right) + C\]

Đặt \[\sqrt x = t \Rightarrow x = {t^2} \Rightarrow dx = 2tdt \Rightarrow I = 2\smallint t\cos tdt.\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = t}\\{dv = costdt}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dt}\\{v = sint}\end{array}} \right.\)

\[ \Rightarrow I = 2\left( {t\sin t - \smallint sintdt + C} \right) = 2\left( {t\sin t + \cos t + C} \right)\]

\[ = 2\left( {\sqrt x \sin \sqrt x + \cos \sqrt x } \right) + C.\]

\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = \frac{1}{{co{s^2}x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = tanx}\end{array}} \right.\)

\[ \Rightarrow F(x) = xtanx - \smallint tanxdx + C = xtanx - \smallint \frac{{sinx}}{{cosx}}dx + C\]

\[ = xtanx + \smallint \frac{{d(cosx)}}{{cosx}} + C = xtanx + ln|cosx| + C.\]

\[ \Rightarrow F(0) = C = 0 \Rightarrow F(\pi ) = 0\]

\[I = \smallint x{\tan ^2}xdx = \smallint x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx = \smallint x.\frac{1}{{{{\cos }^2}x}}dx - \smallint xdx = {I_1} - {I_2}\]

Ta có:\[{I_2} = \smallint xdx = \frac{{{x^2}}}{2} + {C_2},{I_1} = \smallint x\frac{1}{{{{\cos }^2}x}}dx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = \frac{1}{{co{s^2}x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = tanx}\end{array}} \right.\)

\[\begin{array}{*{20}{l}}{ \Rightarrow {I_1} = x\tan x - \smallint \tan xdx + {C_1} = x\tan x - \smallint \frac{{\sin x}}{{\cos x}}dx + {C_1}}\\{ = x\tan x + \smallint \frac{{d\left( {\cos x} \right)}}{{\cos x}} + {C_1} = x\tan x + \ln \left| {\cos x} \right| + {C_1}.}\\{ \Rightarrow I = x\tan x + \ln \left| {\cos x} \right| + {C_1} - \frac{{{x^2}}}{2} - {C_2} = x\tan x + \ln \left| {\cos x} \right| - \frac{{{x^2}}}{2} + C.}\end{array}\]

Ta có:

\[\begin{array}{*{20}{l}}{\cos 2x\ln \left( {\sin x + \cos x} \right) = \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)\ln \left( {\sin x + \cos x} \right)}\\{ \Rightarrow I = \smallint \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)\ln \left( {\sin x + \cos x} \right)dx}\end{array}\]

Đặt\[t = \sin x + \cos x \Rightarrow dt = \left( {\cos x - \sin x} \right)dx\] khi đó ta có:\[I = \smallint t\ln tdt\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = lnt}\\{dv = tdt}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{1}{t}dt}\\{v = \frac{{{t^2}}}{2}}\end{array}} \right.\)

\[ \Rightarrow I = \frac{1}{2}{t^2}lnt - \frac{1}{2}\smallint tdt + C = \frac{1}{2}{t^2}lnt - \frac{{{t^2}}}{4} + {C_1}\]

\[ = \frac{1}{2}{(sinx + cosx)^2}ln(sinx + cosx) - \frac{{{{(sinx + cosx)}^2}}}{4} + {C_1}\]

\[ = \frac{1}{2}(si{n^2}x + co{s^2}x + sin2x)ln(sinx + cosx) - \frac{{1 + sin2x}}{4} + {C_1}\]

\[ = \frac{1}{4}(1 + sin2x)ln{(sinx + cosx)^2} - \frac{{sin2x}}{4} - \frac{1}{4} + {C_1}\]

\[ = \frac{1}{4}(1 + sin2x)ln(1 + sin2x) - \frac{{sin2x}}{4} + C.\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = ln(x + \sqrt {{x^2} + 1} )}\\{dv = dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{{1 + \frac{x}{{\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }}dx}\\{v = x}\end{array}} \right.\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{{\frac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }}}\\{v = x}\end{array}} \right.dx = \frac{{dx}}{{\sqrt {{x^2} + 1} }}\)

\[ \Rightarrow I = x\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \smallint \frac{x}{{\sqrt {{x^2} + 1} }}dx + {C_1}.\]

Đặt\[t = \sqrt {{x^2} + 1} \Rightarrow {t^2} = {x^2} + 1 \Leftrightarrow tdt = xdx\]

\[ \Rightarrow \smallint \frac{x}{{\sqrt {{x^2} + 1} }}dx = \smallint \frac{{tdt}}{t} = \smallint dt = t + {C_2} = \sqrt {{x^2} + 1} + {C_2}\]

Khi đó ta có:\[ \Rightarrow I = x\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \sqrt {{x^2} + 1} + C.\]

Đặt

\(\left\{ {\begin{array}{*{20}{c}}{u = {e^{2x}}}\\{dv = cos3xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = 2{e^{2x}}dx}\\{v = \frac{{sin3x}}{3}}\end{array}} \right. \Rightarrow I = \frac{1}{3}{e^{2x}}sin3x - \frac{2}{3}\smallint {e^{2x}}sin3xdx + {C_1}.\)

Xét nguyên hàm\[\smallint {e^{2x}}\sin 3xdx\] đặt

\(\left\{ {\begin{array}{*{20}{c}}{a = {e^{2x}}}\\{db = sin3xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{da = 2{e^{2x}}}\\{b = - \frac{{cos3x}}{3}}\end{array}} \right.\)

\[ \Rightarrow \smallint {e^{2x}}\sin 3xdx = - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}\smallint {e^{2x}}\cos 3x + {C_1} = - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}I + {C_2}\]

Do đó ta có 

\[\begin{array}{*{20}{l}}{I = \frac{1}{3}{e^{2x}}\sin 3x - \frac{2}{3}\left( { - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}I + {C_2}} \right) + {C_1}}\\{ \Leftrightarrow \frac{{13}}{9}I = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x + C}\\{ \Leftrightarrow I = \frac{1}{{13}}{e^{2x}}\left( {3\sin 3x + 2\cos 3x} \right) + C.}\end{array}\]

Ta có:

\[I = \smallint \frac{{\left( {{x^2} + x} \right){e^x}}}{{x + {e^{ - x}}}}dx = \smallint \frac{{\left( {{x^2} + x} \right){e^x}}}{{\frac{{x{e^x} + 1}}{{{e^x}}}}}dx = \smallint \frac{{\left( {{x^2} + x} \right){e^{2x}}}}{{x{e^x} + 1}}dx = \smallint \frac{{x{e^x}\left( {x + 1} \right){e^x}}}{{x{e^x} + 1}}dx.\]

Đặt

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{u = x{e^x}}\\{dv = \frac{{(x + 1){e^x}}}{{x{e^x} + 1}}dx = \frac{{d(x{e^x} + 1)}}{{x{e^x} + 1}}}\end{array}} \right.\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = ({e^x} + x{e^x})dx = (x + 1){e^x}dx}\\{v = ln|x{e^x} + 1|}\end{array}} \right.\end{array}\)

Khi đó ta có: \[I = x{e^x}\ln \left| {x{e^x} + 1} \right| - \smallint \ln \left| {x{e^x} + 1} \right|\left( {x + 1} \right){e^x}dx + C.\]

Đặt\[t = x{e^x} + 1 \Rightarrow dt = \left( {{e^x} + x{e^x}} \right)dx = \left( {x + 1} \right){e^x}dx\]

\[ \Rightarrow \smallint \ln \left| {x{e^x} + 1} \right|\left( {x + 1} \right){e^x}dx = \smallint \ln \left| t \right|dt\]

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = ln|t|}\\{dv = dt}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{1}{t}dt}\\{v = t}\end{array}} \right.\)

\[ \Rightarrow \smallint \ln \left| t \right|dt = \ln \left| t \right|.t - \smallint dt + C = \ln \left| t \right|.t - t + C\]

\[ = \left( {x{e^x} + 1} \right)\ln \left| {x{e^x} + 1} \right| - \left( {x{e^x} + 1} \right) + C.\]

Vậy\[I = x{e^x}\ln \left| {x{e^x} + 1} \right| - \left( {x{e^x} + 1} \right)\ln \left| {x{e^x} + 1} \right| + \left( {x{e^x} + 1} \right) + C\]

\[ = x{e^x} + 1 - \ln \left| {x{e^x} + 1} \right| + C.\]

Ta có:\[\frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} - \frac{1}{{{x^2} + 1}}\]

\[ \Rightarrow \smallint \frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = \smallint \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx - \smallint \frac{1}{{{x^2} + 1}}dx\,\,\left( 1 \right)\]

Ta tính\[\smallint \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = \smallint \frac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\] bằng phương pháp tích phân từng phân như sau:

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = \frac{{d({x^2} + 1)}}{{{{({x^2} + 1)}^2}}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - \frac{1}{{{x^2} + 1}}}\end{array}} \right.\)

\[ \Rightarrow \smallint \frac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = - \frac{x}{{{x^2} + 1}} + \smallint \frac{{dx}}{{{x^2} + 1}} + C\,\,\left( 2 \right)\]

Từ (1) và (2) suy ra 

\[\smallint \frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = - \frac{x}{{{x^2} + 1}} + \smallint \frac{{dx}}{{{x^2} + 1}} + C - \smallint \frac{1}{{{x^2} + 1}}dx = - \frac{x}{{{x^2} + 1}} + C.\]

Vì \[x{e^x}\] là một nguyên hàm của hàm số f(−x) nên

\[{\left( {x{e^x}} \right)^\prime } = f\left( { - x} \right) \Leftrightarrow f\left( { - x} \right) = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right)\]

\[ \Rightarrow f\left( x \right) = {e^{ - x}}\left( {1 - x} \right)\]

\[\begin{array}{l} \Rightarrow f\prime (x) = - {e^{ - x}}(1 - x) - {e^{ - x}} = - e - x(2 - x) = (x - 2){e^{ - x}}\\ \Rightarrow f\prime (x){e^x} = (x - 2){e^{ - x}}.{e^x} = x - 2\\ \Rightarrow F(x) = \smallint f(x)dx = \smallint (x - 2)dx = \frac{{{x^2}}}{2} - 2x + C\\F(0) = 1 \Rightarrow C = 1 \Rightarrow F(x) = \frac{{{x^2}}}{2} - 2x + 1\\ \Rightarrow F( - 1) = \frac{{{{( - 1)}^2}}}{2} - 2( - 1) + 1 = \frac{7}{2}\end{array}\]

Ta có:

\[\begin{array}{l}F\prime (x) = f(x)\\ \Leftrightarrow f\prime (x) - {e^x} - 1 = f(x)\\ \Leftrightarrow f\prime (x) - f(x) = {e^x} + 1\\ \Leftrightarrow {e^{ - x}}.f\prime (x) - {e^{ - x}}.f(x) = 1 + {e^{ - x}}\\ \Leftrightarrow [{e^{ - x}}.f(x)]\prime = 1 + {e^{ - x}}\\ \Leftrightarrow \smallint [{e^{ - x}}.f(x)]\prime dx = \smallint (1 + {e^{ - x}})dx\\ \Leftrightarrow {e^{ - x}}.f(x) = x - {e^{ - x}} + C\\ \Leftrightarrow f(x) = x.{e^x} - 1 + C.{e^x}\end{array}\]

Thay x=0 ta có: \[f\left( 0 \right) = - 1 + C = 1 \Leftrightarrow C = 2\]

\[\begin{array}{l} \Rightarrow f(x) = x.{e^x} - 1 + 2{e^x} = (x + 2){e^x} - 1\\ \Rightarrow \smallint f(x)dx = \smallint [(x + 2){e^x} - 1]dx\\ = \smallint (x + 2){e^x}dx - \smallint dx\\ = (x + 2){e^x} - \smallint {e^x}dx - x + C\\ = (x + 2){e^x} - {e^x} - x + C\\ = (x + 1){e^x} - x + C\end{array}\]

Vì\[F\left( x \right) = {\left[ {f\left( x \right)} \right]^{2020}}\]  là một nguyên hàm của\[2020x.{e^x}\] nên

\[\begin{array}{*{20}{l}}{F'\left( x \right) = 2020x.{e^x}}\\{ \Leftrightarrow 2020{f^{2019}}\left( x \right).f'\left( x \right) = 2020x.{e^x}}\\{ \Leftrightarrow {f^{2019}}\left( x \right).f'\left( x \right) = x.{e^x}}\end{array}\]

Lấy nguyên hàm hai vế ta được:

\[\begin{array}{*{20}{l}}{\smallint {f^{2019}}\left( x \right).f'\left( x \right)dx = \smallint x.{e^x}dx}\\{ \Leftrightarrow \smallint {f^{2019}}\left( x \right)d\left[ {f\left( x \right)} \right] = x.{e^x} - \smallint {e^x}dx}\\{ \Leftrightarrow \frac{{{f^{2020}}\left( x \right)}}{{2020}} = x.{e^x} - {e^x} + C}\\{ \Leftrightarrow {f^{2020}}\left( x \right) = 2020\left( {x - 1} \right){e^x} + 2020C}\end{array}\]

Có \[f\left( 1 \right) = 1 \Leftrightarrow 0 = 2020C \Leftrightarrow C = 0\] do đó\[{f^{2020}}\left( x \right) = 2020\left( {x - 1} \right){e^x}\]

\[ \Rightarrow I = \smallint {f^{2020}}\left( x \right)dx = \smallint 2020\left( {x - 1} \right){e^x}dx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x - 1}\\{dv = {e^x}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = {e^x}}\end{array}} \right.\)

Khi đó

\[\begin{array}{*{20}{l}}{\,\,\,\,\,\,I = 2020\left[ {\left( {x - 1} \right){e^x} - \smallint {e^x}dx + C} \right]}\\{ \Leftrightarrow I = 2020\left[ {\left( {x - 1} \right){e^x} - {e^x} + C} \right]}\\{ \Leftrightarrow I = 2020\left[ {\left( {x - 2} \right){e^x} + C} \right]}\\{ \Leftrightarrow I = 2020\left( {x - 2} \right){e^x} + C}\end{array}\]