Sử dụng phương pháp tích phân từng phần để tính tích phân

Đặt $\left\{ \begin{array}{l}u = f\left( x \right)\\{\rm{d}}v = g'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = f'\left( x \right){\rm{d}}x\\v = g\left( x \right)\end{array} \right.,$ khi đó $I = \left. {f\left( x \right).g\left( x \right)} \right|_a^b - \int\limits_a^b {f'\left( x \right).g\left( x \right){\rm{d}}x} .$

Đặt $\left\{ \begin{array}{l}u = {x^2}\\{\rm{d}}v = \cos x\,{\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = 2x\,{\rm{d}}x\\v = \sin x\end{array} \right.,$ khi đó $I = \left. {{x^2}\sin x} \right|_0^{\dfrac{\pi }{2}} - 2\int\limits_0^{\dfrac{\pi }{2}} {x\sin x\,{\rm{d}}x} .$

Đặt $\left\{ \begin{array}{l}u = g\left( x \right)\\{\rm{d}}v = f'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = g'\left( x \right){\rm{d}}x\\v = f\left( x \right)\end{array} \right..$

Khi đó $\int\limits_0^1 {g\left( x \right).f'\left( x \right){\rm{d}}x}  = \left. {\left[ {g\left( x \right).f\left( x \right)} \right]} \right|_0^1 - \int\limits_0^1 {g'\left( x \right).f\left( x \right){\rm{d}}x}  \Leftrightarrow \left. {\left[ {g\left( x \right).f\left( x \right)} \right]} \right|_0^1 = 3.$

Mặt khác $I = \int\limits_0^1 {{{\left[ {f\left( x \right).g\left( x \right)} \right]}^\prime }\,{\rm{d}}x}  = \left. {\left[ {f\left( x \right).g\left( x \right)} \right]} \right|_0^1\,\, \Rightarrow \,\,I = 3.$

Vì ${x^2}$ là một nguyên hàm của hàm số $f\left( x \right){e^{2x}} \Rightarrow \int {f\left( x \right){e^{2x}}\,{\rm{d}}x}  = {x^2}.$

Đặt $\left\{ \begin{array}{l}u = {e^{2x}}\\{\rm{d}}v = f'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = 2{e^{2x}}{\rm{d}}x\\v = f\left( x \right)\end{array} \right.,$ khi đó $\int\limits_0^1 {f'\left( x \right){e^{2x}}{\rm{d}}x}  = \left. {f\left( x \right){e^{2x}}} \right|_0^1 - 2\int\limits_0^1 {f\left( x \right){e^{2x}}\,{\rm{d}}x} .$

Suy ra $I = {e^2}f\left( 1 \right) - f\left( 0 \right) - 2\left. {{x^2}} \right|_0^1 =  2-0 - 2 =  0$

Vậy $I =  0$

Đặt $\left\{ \begin{array}{l}u = x + \ln x\\{\rm{d}}v = \dfrac{{{\rm{d}}x}}{{{{\left( {x + 1} \right)}^3}}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{x + 1}}{x}\,{\rm{d}}x\\v =  - \dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}\end{array} \right.$.

Khi đó $I = \left. { - \dfrac{{x + \ln x}}{{2{{\left( {x + 1} \right)}^2}}}} \right|_1^2 + \int\limits_1^2 {\dfrac{{x + 1}}{x}.\dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}{\rm{d}}x} .$

$ =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\dfrac{{{\rm{d}}x}}{{x\left( {x + 1} \right)}}} $$=  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right){\rm{d}}x} .$

$\begin{array}{l} =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\left. {\left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_1^2\\ = \dfrac{1}{{72}} - \dfrac{1}{{18}}\ln 2 + \dfrac{1}{2}\left( {\ln 2 - \ln 3 + \ln 2} \right)\\ = \dfrac{1}{{72}} + \dfrac{{17}}{{18}}\ln 2 - \dfrac{1}{2}\ln 3\\ = a + b.\ln 2 - c.\ln 3.\end{array}$

Vậy $\left\{ \begin{array}{l}a = \dfrac{1}{{72}}\\b = \dfrac{{17}}{{18}}\\c = \dfrac{1}{2}\end{array} \right. \Rightarrow \,\dfrac{c}{a} = \dfrac{1}{2}:\dfrac{1}{{72}} = 36.$

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

$I = {x^2}\left( {1 - \ln x} \right)\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. - \int\limits_1^e { - xdx}  =  - 1 + \dfrac{{{x^2}}}{2}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. =  - 1 + \left( {\dfrac{{{e^2}}}{2} - \dfrac{1}{2}} \right) = \dfrac{{{e^2} - 3}}{2}$

Dùng máy tính kiểm tra từng đáp án hoặc:

Đặt $u = \ln x,dv = xdx \Rightarrow du = \dfrac{{dx}}{x},v = \dfrac{{{x^2}}}{2}$

$I = \dfrac{{{x^2}\ln x}}{2}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. - \int\limits_1^e {\dfrac{x}{2}dx}  = \dfrac{{{e^2}}}{2} - \dfrac{{{x^2}}}{4}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. = \dfrac{{{e^2}}}{2} - \left( {\dfrac{{{e^2}}}{4} - \dfrac{1}{4}} \right) = \dfrac{{{e^2} + 1}}{4}$

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

$\begin{array}{l} \Rightarrow I =  - \dfrac{{\ln x}}{{x + 1}}\left| {_1^{{2^{1000}}} + \int\limits_1^{{2^{1000}}} {\dfrac{1}{{x + 1}}.} \dfrac{{dx}}{x} = } \right. - \dfrac{{\ln {2^{1000}}}}{{{2^{1000}} + 1}} + \int\limits_1^{{2^{1000}}} {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right)dx}  \\ =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \left. {\ln \left| {\dfrac{x}{{x + 1}}} \right|} \right|_1^{{2^{1000}}} =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1000}}}}{{{2^{1000}} + 1}} - \ln \dfrac{1}{2} \\ =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1001}}}}{{{2^{1000}} + 1}}\end{array}$

Đặt$f\left( x \right) = {e^{2x}}\left( {a\cos 3x + b\sin 3x} \right) + c$. Ta có

$f'\left( x \right) = 2a{e^{2x}}\cos 3x - 3a{e^{2x}}\sin 3x + 2b{e^{2x}}\sin 3x + 3b{e^{2x}}\cos 3x $

$= \left( {2a + 3b} \right){e^{2x}}\cos 3x + \left( {2b - 3a} \right){e^{2x}}\sin 3x$

Để $f\left( x \right)$ là một nguyên hàm của hàm số ${e^{2x}}\cos 3x$, điều kiện là

$f'\left( x \right) = {e^{2x}}\cos 3x \Leftrightarrow \left\{ \begin{array}{l}2a + 3b = 1\\2b - 3a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{2}{{13}}\\b = \dfrac{3}{{13}}\end{array} \right. \Rightarrow a + b = \dfrac{5}{{13}}$

Đặt $u=x+1;dv=f'(x)dx$ thì $du=dx;v=f(x)$

Ta có:

$\begin{array}{l}\int\limits_0^1 {(x + 1)f'(x)d{\rm{x}} = 10}  \Leftrightarrow \left. {(x + 1)f(x)} \right|_0^1 - \int\limits_0^1 {f(x)d{\rm{x}} = 10 = } 2f(1) - f(0) - \int\limits_0^1 {f(x)d{\rm{x}}} \\ \to \int\limits_0^1 {f(x)d{\rm{x}}}  =  - 8.\end{array}$

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

$\Rightarrow \int {f(x)\sin {\rm{x}}dx =  - f(x).\cos x + \int {f'(x).\cos xdx} }$

Nên suy ra \(f'(x) = {\pi ^x} \Rightarrow f(x) = \int {{\pi ^x}} dx = \dfrac{{{\pi ^x}}}{{\ln \pi }}\).

$\begin{array}{l}u\left( x \right) = x \Rightarrow u'\left( x \right) = 1\\v'\left( x \right) = \cos 2x \Rightarrow v\left( x \right) = \dfrac{{\sin 2x}}{2}\\ \Rightarrow \int\limits_0^1 {x\cos 2xdx = \left. {\dfrac{x}{2}\sin 2x} \right|_0^1}  - \dfrac{1}{2}\int\limits_0^1 {\sin 2xdx}  = \left. {\dfrac{x}{2}\sin 2x} \right|_0^1 + \left. {\dfrac{{\cos 2x}}{4}} \right|_0^1\\ = \dfrac{1}{2}\sin 2 + \dfrac{1}{4}\cos 2 - \dfrac{1}{4} = \dfrac{1}{4}\left( {2\sin 2 + \cos 2 - 1} \right)\\ \Rightarrow a = 2;b = 1;c =  - 1\end{array}$

Khi đó $a - b + c = 2 - 1 - 1 = 0$

$I = \int\limits_0^\pi  {{x^2}\cos xdx} $

Đặt $\left\{ \begin{array}{l}u = {x^2}\\dv = \cos xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = 2xdx\\v = \int {\cos xdx} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = 2xdx\\v = \sin x\end{array} \right.$

$ \Rightarrow I = \left. {{x^2}.\sin x} \right|_0^\pi  - 2\int\limits_0^\pi  {x.\sin xdx} $ 

Đặt $\left\{ \begin{array}{l}u = \ln {(2x + 1)^{2017}}\\dv = xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{2017.2.{{(2x + 1)}^{2016}}}}{{{{(2x + 1)}^{2017}}}}dx = \dfrac{{4034}}{{2x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.$

$\left. {I = \ln {{(2x + 1)}^{2017}}.\dfrac{{{x^2}}}{2}} \right|_0^4 - \int_0^4 {\dfrac{{{x^2}}}{2}.\dfrac{{4034}}{{2x + 1}}dx}   $

$= \ln {(2.4 + 1)^{2017}}.\dfrac{{{4^2}}}{2} - 0 - 2017\int_0^4 {\dfrac{{{x^2}}}{{2x + 1}}dx}$

$ = 8\ln {9^{2017}} - 2017\int_0^4 {(\dfrac{1}{2}x - \dfrac{1}{4} + \dfrac{{\dfrac{1}{4}}}{{2{\rm{x}} + 1}})dx}   $

$= 8\ln {9^{2017}} - \dfrac{{2017}}{2}.\left. {\dfrac{{{x^2}}}{2}} \right|_0^4 + \dfrac{{2017}}{4}\left. x \right|_0^4 - \dfrac{{2017}}{4}\int_0^4 {\dfrac{1}{2}.\dfrac{1}{{2x + 1}}d(2x + 1)}$

$\begin{array}{l} = 8\ln {9^{2017}} - \dfrac{{2017}}{4}{.4^2} + \dfrac{{2017}}{4}.4 - \dfrac{{2017}}{8}\left. {\ln \left| {2x + 1} \right|} \right|_0^4 \\ = 8\ln {9^{2017}} - 6051 - \dfrac{{2017}}{8}.(\ln 9 - \ln 1)\\ = 8\ln {9^{2017}} - 6051 - \dfrac{{2017}}{8}.\ln 9 = \dfrac{{127071}}{4}.\ln 3 - 6051\end{array}$

$ \Rightarrow b + c = 127075$

\(I = \int_1^n {\ln xdx} \)

Đặt $\ln x = u;dv=dx$ . Suy ra \(\dfrac{1}{x}dx = du; v = x\)

\(I = \left. {\left( {x\ln x} \right)} \right|_1^n - \int_1^n {\dfrac{x}{x}dx = n\ln n - n + 1} \)

Biểu thức ban đầu sẽ là: $n - 1$

Để $n-1\le 2017$ thì \(n \le 2018\) và $n$ nguyên dương

Nên sẽ có $2018$  giá trị của $n$

Đặt : \(\left\{ \begin{array}{l}u = x\\dv = \cos 2xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dx\\v = \dfrac{1}{2}.\sin 2x\end{array} \right.\)

Suy ra: $\int\limits_0^{\dfrac{\pi }{4}} {x.\cos xdx}  = \left. {\left( {x.\dfrac{1}{2}.{\mathop{\rm s}\nolimits} {\rm{in2x}}} \right)} \right|_0^{\dfrac{\pi }{4}} - \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {\sin 2xdx}  $

$= \dfrac{\pi }{8} + \left. {\dfrac{1}{4}\cos 2x} \right|_0^{\dfrac{\pi }{4}} =  - \dfrac{1}{4} + \dfrac{\pi }{8}$

\( \Rightarrow a =  - \dfrac{1}{4};b = \dfrac{1}{8} \Rightarrow S = a + 2b = 0\)

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

$\Rightarrow I = \left. {\dfrac{{x{e^{2x}}}}{2}} \right|_0^1 - \int\limits_0^1 {\dfrac{{{e^{2x}}}}{2}dx = } \left. {\left( {\dfrac{{x{e^{2x}}}}{2} - \dfrac{{{e^{2x}}}}{4}} \right)} \right|_0^1 = \dfrac{{{e^2}}}{4} + \dfrac{1}{4}$

\( \Rightarrow a = \dfrac{1}{4};b = \dfrac{1}{4} \Rightarrow a + b = \dfrac{1}{2}\) 

Đặt \(\left\{ \begin{array}{l}u = {e^x}\\dv = \sin xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = {e^x}dx\\v =  - \cos x\end{array} \right.\) \(I = \int\limits_0^{\dfrac{\pi }{2}} {{e^x}\sin xdx}  = \left. { - {e^x}\cos x} \right|_0^{\dfrac{\pi }{2}} + \int\limits_0^{\dfrac{\pi }{2}} {{e^x}\cos xdx}  = 1 + \int\limits_0^{\dfrac{\pi }{2}} {{e^x}\cos xdx} \)

Đặt \(\left\{ \begin{array}{l}u = {e^x}\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = {e^x}dx\\v = \sin xdx\end{array} \right.\)

Khi đó \(\int\limits_0^{\dfrac{\pi }{2}} {{e^x}\cos xdx}  = \left. {{e^x}\sin x} \right|_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {{e^x}\sin xdx}  = {e^{\dfrac{\pi }{2}}} - \int\limits_0^{\dfrac{\pi }{2}} {{e^x}\sin xdx}  = {e^{\dfrac{\pi }{2}}} - I\)

Do đó \(I = 1 + {e^{\dfrac{\pi }{2}}} - I \Leftrightarrow 2I = {e^{\dfrac{\pi }{2}}} + 1 \Leftrightarrow I = \dfrac{{{e^{\dfrac{\pi }{2}}} + 1}}{2} \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 2\end{array} \right.\)

Quan sát các đáp án ta thấy đáp án A thỏa mãn.

\(I = \int\limits_0^1 {\left( {x + \sqrt {{x^2} + 15} } \right)dx}  = \int\limits_0^1 {xdx}  + \int\limits_0^1 {\sqrt {{x^2} + 15} dx} \)

\({I_1} = \int\limits_0^1 {xdx}  = \left. {\dfrac{1}{2}{x^2}} \right|_0^1 = \dfrac{1}{2}\)

\(\begin{array}{l}{I_2} = \int\limits_0^1 {\sqrt {{x^2} + 15} dx}  = \left. {x\sqrt {{x^2} + 15} } \right|_0^1 - \int\limits_0^1 {x.\dfrac{x}{{\sqrt {{x^2} + 15} }}dx} \\\,\,\,\,\,\, = 4 - \int\limits_0^1 {\dfrac{{{x^2}}}{{\sqrt {{x^2} + 15} }}dx}  = 4 - \int\limits_0^1 {\sqrt {{x^2} + 15} dx}  + \int\limits_0^1 {\dfrac{{15}}{{\sqrt {{x^2} + 15} }}dx} \\ \Rightarrow 2{I_2} = 4 + 15\int\limits_0^1 {\dfrac{1}{{\sqrt {{x^2} + 15} }}dx} \end{array}\)

Đặt \(x + \sqrt {{x^2} + 15}  = t \Rightarrow \left( {1 + \dfrac{x}{{\sqrt {{x^2} + 15} }}} \right)dx = dt \Leftrightarrow \dfrac{{dx}}{{\sqrt {{x^2} + 15} }} = \dfrac{{dt}}{t}\).

Khi đó: \(\int\limits_0^1 {\dfrac{1}{{\sqrt {{x^2} + 15} }}dx}  = \int\limits_{\sqrt {15} }^5 {\dfrac{{dt}}{t}}  = \left. {\ln \left| t \right|} \right|_{\sqrt {15} }^5 = \ln 5 - \ln \sqrt {15}  = \dfrac{1}{2}\ln 5 - \dfrac{1}{2}\ln 3\)

\( \Rightarrow 2{I_2} = 4 + 15.\left( {\dfrac{1}{2}\ln 5 - \dfrac{1}{2}\ln 3} \right) \Leftrightarrow {I_2} = 2 + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3\)

\(I = {I_1} + {I_2} = \dfrac{1}{2} + 2 + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3 = \dfrac{5}{2} + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3 \Rightarrow a + b + c = \dfrac{5}{2} + \dfrac{{15}}{4} - \dfrac{{15}}{4} = \dfrac{5}{2}\)

Vì \(f\left( x \right)\) là hàm số chẵn và liên tục trên \(\left[ { - 1;1} \right]\) nên \(\int\limits_{ - 1}^1 {f\left( x \right)dx}  = 2\int\limits_0^1 {f\left( x \right)dx}  = \dfrac{{86}}{{15}}\) \( \Rightarrow \int\limits_0^1 {f\left( x \right)dx}  = \dfrac{{43}}{{15}}\).

Xét tích phân \(I = \int\limits_0^1 {xf'\left( x \right)dx} \).

Đặt \(\left\{ \begin{array}{l}u = x\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = f\left( x \right)\end{array} \right.\), khi đó ta có:

\(I = \left. {xf\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)dx}  = f\left( 1 \right) - \int\limits_0^1 {f\left( x \right)dx}  = 5 - \dfrac{{43}}{{15}} = \dfrac{{32}}{{15}}\).