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

Xét tích phân \(I = \int\limits_0^{{\pi ^2}} {f\left( {\sqrt x } \right)\cos \left( {\sqrt x } \right)dx} \) .

Đặt \(t = \sqrt x  \Rightarrow {t^2} = x \Rightarrow 2tdt = dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = {\pi ^2} \Rightarrow t = \pi \end{array} \right.\), khi đó ta có \(I = \int\limits_0^\pi  {f\left( t \right)\cos \left( t \right)2tdt}  = \int\limits_0^\pi  {2f\left( x \right)\cos x.xdx} \).

Xét tích phân  \(\int\limits_0^\pi  {xf'\left( x \right)\sin xdx}  = 5\).

Đặt \(\left\{ \begin{array}{l}u = x\sin x\\f'\left( x \right)dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = \left( {\sin x + x\cos x} \right)du\\v = f\left( x \right)\end{array} \right.\).

\(\begin{array}{l} \Rightarrow \int\limits_0^\pi  {xf'\left( x \right)\sin xdx}  = 5\\ \Leftrightarrow \left. {\left( {x\sin x.f\left( x \right)} \right)} \right|_0^\pi  - \int\limits_0^\pi  {\left[ {f\left( x \right)\sin x + xf\left( x \right)\cos x} \right]dx}  = 5\\ \Leftrightarrow  - \int\limits_0^\pi  {f\left( x \right)\sin xdx}  - \int\limits_0^\pi  {xf\left( x \right)\cos xdx}  = 5\\ \Leftrightarrow  - 20 - \dfrac{I}{2} = 5\\ \Leftrightarrow \dfrac{I}{2} =  - 25\\ \Leftrightarrow I =  - 50\end{array}\)

Ta có: \(\int\limits_1^3 {\left( {4 - x} \right)f\left( x \right)dx}  = 4\int\limits_1^3 {f\left( x \right)dx}  - \int\limits_1^3 {xf\left( x \right)dx} \).

Đặt \(t = 4 - x \Rightarrow dt =  - dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 3\\x = 3 \Rightarrow t = 1\end{array} \right.\), khi đó ta có:

\(\int\limits_1^3 {\left( {4 - x} \right)f\left( x \right)dx}  =  - \int\limits_3^1 {tf\left( {4 - t} \right)dt} \) \( = \int\limits_1^3 {tf\left( {4 - t} \right)dt}  = \int\limits_1^3 {tf\left( t \right)dt} \)\( = \int\limits_1^3 {xf\left( x \right)dx} \).

\(\begin{array}{l} \Rightarrow \int\limits_1^3 {xf\left( x \right)dx}  = 4\int\limits_1^3 {f\left( x \right)dx}  - \int\limits_1^3 {xf\left( x \right)dx} \\ \Leftrightarrow 2\int\limits_1^3 {f\left( x \right)dx}  = \int\limits_1^3 {xf\left( x \right)dx}  =  { - 2} \end{array}\)

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

Khi đó ta có:

\(\begin{array}{l}\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right)dx} \\ = \left. {\sin x.f\left( x \right)} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {x{{\sin }^2}xdx} \\ = \sin \dfrac{\pi }{2}.f\left( {\dfrac{\pi }{2}} \right) - \int\limits_0^{\frac{\pi }{2}} {x\dfrac{{1 - \cos 2x}}{2}dx} \\ = 2 - \dfrac{1}{2}\left( {\int\limits_0^{\frac{\pi }{2}} {xdx}  - \int\limits_0^{\frac{\pi }{2}} {x\cos 2xdx} } \right)\\ = 2 - \dfrac{1}{2}\left( {\left. {\dfrac{{{x^2}}}{2}} \right|_0^{\frac{\pi }{2}} - I} \right)\\ = 2 - \dfrac{1}{2}\left( {\dfrac{{{\pi ^2}}}{8} - I} \right)\\ = 2 - \dfrac{{{\pi ^2}}}{{16}} + \dfrac{I}{2}\end{array}\)

Xét tích phân \(I = \int\limits_0^{\frac{\pi }{2}} {x\cos 2xdx} \).

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

\(\begin{array}{l}I = \left. {x.\dfrac{{\sin 2x}}{2}} \right|_0^{\frac{\pi }{2}} - \dfrac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} \\I = \dfrac{\pi }{2}.\dfrac{{\sin \pi }}{2} - 0 + \dfrac{1}{2}.\left. {\dfrac{{\cos 2x}}{2}} \right|_0^{\frac{\pi }{2}}\\I = \dfrac{1}{4}\left( {\cos \pi  - \cos 0} \right)\\I = \dfrac{1}{4}\left( { - 1 - 1} \right) =  - \dfrac{1}{2}\end{array}\)

Do đó \(\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right)dx}  = 2 - \dfrac{{{\pi ^2}}}{{16}} - \dfrac{1}{4} = \dfrac{7}{4} - \dfrac{{{\pi ^2}}}{{16}}\)

\( \Rightarrow a = 7,\,\,b = 4,\,\,c = 16\)

Vậy \(a + b + c = 7 + 4 + 16 = 27\).

Đặt $\left\{ \begin{array}{l}u = \ln \left( {3\sin x + \cos x} \right)\\dv = \dfrac{{dx}}{{{{\sin }^2}x}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = \dfrac{{3\cos x - \sin x}}{{3\sin x + \cos x}}\,\,dx\\v =  - \cot x - 3 =  - \dfrac{{3\sin x + \cos x}}{{\sin x}}\end{array} \right.$.

Khi đó $I = \left. {\left[ { - \left( {\cot x + 3} \right)\ln \left( {3\sin x + \cos x} \right)} \right]} \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} + \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\dfrac{{3\cos x - \sin x}}{{\sin x}}\,\,dx} .$

$\begin{array}{l} = 4.\ln 2\sqrt 2  - 3.\ln 3 - \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {dx}  + 3.\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\dfrac{{d\left( {\sin x} \right)}}{{\sin x}}} \\ = 4.\ln 2\sqrt 2  - 3.\ln 3 - \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {dx}  + 3\left. {\ln \left| {\sin x} \right|} \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}\\ = 4.\ln 2\sqrt 2  - 3.\ln 3 - \dfrac{\pi }{4} - 3.\ln \dfrac{1}{{\sqrt 2 }}\\ = 12ln\sqrt 2  - 3\ln 3 - \dfrac{\pi }{4} + 3\ln \sqrt 2  = 15.\ln \sqrt 2  - 3.\ln 3 - \dfrac{\pi }{4}\\ \Rightarrow \left\{ \begin{array}{l}m = 15\\n =  - 3\end{array} \right. \Rightarrow m + n = 12.\end{array}$

Xét tích phân \(\int\limits_4^7 {f\left( {\dfrac{x}{2}} \right)dx}  = \dfrac{a}{b}\).

Đặt \(t = \dfrac{x}{2} \Rightarrow dt = \dfrac{1}{2}dx \Leftrightarrow dx = 2dt\). Đổi cận: \(\left\{ \begin{array}{l}x = 4 \Rightarrow t = 2\\x = 7 \Rightarrow t = \dfrac{7}{2}\end{array} \right.\).

Khi đó ta có: \(I = 2\int\limits_2^{\dfrac{7}{2}} {f\left( t \right)dt} \).

Đặt \(\left\{ \begin{array}{l}u = f\left( t \right)\\dv = dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = f'\left( t \right)dt\\v = t - \dfrac{7}{2}\end{array} \right.\), khi đó ta có:

\(\begin{array}{l}I = 2\left( {\left. {\left( {t - \dfrac{7}{2}} \right)f\left( t \right)} \right|_2^{\dfrac{7}{2}} - \int\limits_2^{\dfrac{7}{2}} {\left( {t - \dfrac{7}{2}} \right)f'\left( t \right)dt} } \right)\\I = 2\left( {\dfrac{7}{2}f\left( 0 \right) - \int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right)f'\left( x \right)dx} } \right)\\I = 2\left( {\dfrac{7}{2}f\left( 2 \right) - \int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right).\dfrac{{x + 7}}{{\sqrt {2x - 3} }}dx} } \right)\\I =  - 2\int\limits_2^{\dfrac{7}{2}} {\left( {x - \dfrac{7}{2}} \right).\dfrac{{x + 7}}{{\sqrt {2x - 3} }}dx} \\I = \dfrac{{236}}{{15}}\\ \Rightarrow a = 236,\,\,b = 15.\end{array}\)

Vậy \(a + b = 236 + 15 = 251\) .

Xét tích phân: \(I = \int\limits_0^1 {\dfrac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} \).

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

Khi đó ta có:

\(\begin{array}{l}\,\,\,\,\,\,I = \left. {\dfrac{{x - 1}}{{2\left( {x + 1} \right)}}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {\dfrac{{x - 1}}{{2\left( {x + 1} \right)}}f'\left( x \right)dx} \\ \Leftrightarrow I = \dfrac{1}{2}f\left( 0 \right) - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{x - 1}}{{x + 1}}f'\left( x \right)dx} \\ \Leftrightarrow 4\ln 2 - 2 = \dfrac{1}{2}.2 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{x - 1}}{{x + 1}}f'\left( x \right)dx} \\ \Leftrightarrow \int\limits_0^1 {\dfrac{{x - 1}}{{x + 1}}f'\left( x \right)dx}  = 6 - 8\ln 2\end{array}\)

Xét \(\int\limits_0^1 {{{\left( {f'\left( x \right) + k\dfrac{{x - 1}}{{x + 1}}} \right)}^2}dx}  = 0\).

\(\begin{array}{l} \Leftrightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx}  + 2k\int\limits_0^1 {\dfrac{{x - 1}}{{x + 1}}f'\left( x \right)dx}  + {k^2}\int\limits_0^1 {{{\left( {\dfrac{{x - 1}}{{x + 1}}} \right)}^2}dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\int\limits_0^1 {{{\left( {1 - \dfrac{2}{{x + 1}}} \right)}^2}dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\int\limits_0^1 {\left( {1 - \dfrac{4}{{x + 1}} + \dfrac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\left. {\left( {x - 4\ln \left| {x + 1} \right| - \dfrac{4}{{x + 1}}} \right)} \right|_0^1 = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\left( {1 - 4\ln 2 - 2 + 4} \right) = 0\\ \Leftrightarrow \left( {3 - 4\ln 2} \right){k^2} - 4\left( {3 - 4\ln 2} \right)k + 4\left( {3 - 4\ln 2} \right) = 0\\ \Leftrightarrow {k^2} - 4k + 4 = 0 \Leftrightarrow {\left( {k - 2} \right)^2} = 0 \Leftrightarrow k = 2\end{array}\)

Khi đó ta có \(\int\limits_0^1 {{{\left( {f'\left( x \right) - 2.\dfrac{{x - 1}}{{x + 1}}} \right)}^2}dx}  = 0 \Leftrightarrow f'\left( x \right) = 2.\dfrac{{x - 1}}{{x + 1}}\).

\( \Rightarrow f\left( x \right) = \int {f'\left( x \right)dx}  = 2\int {\dfrac{{x - 1}}{{x + 1}}dx} \) \( = 2\int {\left( {1 - \dfrac{2}{{x + 1}}} \right)dx}  = 2\left( {x - 2\ln \left| {x + 1} \right|} \right) + C\).

Có \(f\left( 0 \right) = 2\) \( \Rightarrow 2\left( {0 - 2\ln 1} \right) + C = 2 \Leftrightarrow C = 2\).

\( \Rightarrow f\left( x \right) = 2\left( {x - 2\ln \left| {x + 1} \right|} \right) + 2 = 2x - 4\ln \left| {x + 1} \right| + 2\).

\(\begin{array}{l} \Rightarrow \int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\left[ {2x - 4\ln \left| {x + 1} \right| + 2} \right]dx} \\ = \left. {\left( {{x^2} + 2x} \right)} \right|_0^1 - 4\int\limits_0^1 {\ln \left| {x + 1} \right|dx}  = 3 - 4J\end{array}\).

Ta có: \(J = \int\limits_0^1 {\ln \left| {x + 1} \right|dx}  = \int\limits_0^1 {\ln \left( {x + 1} \right)dx} \).

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

\(\begin{array}{l} \Rightarrow J = \left. {\left( {x + 1} \right)\ln \left( {x + 1} \right)} \right|_0^1 - \int\limits_0^1 {dx} \\ \Rightarrow J = 2\ln 2 - 1.\ln 1 - \left. x \right|_0^1\\ \Rightarrow J = 2\ln 2 - 1\end{array}\)

Vậy \(\int\limits_0^1 {f\left( x \right)dx}  = 3 - 4\left( {2\ln 2 - 1} \right) = 7 - 8\ln 2\).

Ta có : \(\left( {x\sin x + \cos x} \right)' = \sin x + x\cos x - \sin x = x\cos x\)

$ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}{\rm{d}}x}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\dfrac{x}{{\cos x}}.x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dv} $

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

Khi đó

$\begin{array}{l}I = \left. { - \dfrac{x}{{\cos x}}.\dfrac{1}{{x\sin x + \cos x}}} \right|_0^{\dfrac{\pi }{4}} + \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{\rm{d}}x}}{{{{\cos }^2}x}}}  = \\ = \dfrac{{ - \dfrac{\pi }{4}}}{{\dfrac{{\sqrt 2 }}{2}}}.\dfrac{1}{{\dfrac{\pi }{4}\dfrac{{\sqrt 2 }}{2} + \dfrac{{\sqrt 2 }}{2}}} + \left. {\tan x} \right|_0^{\dfrac{\pi }{4}}\\ = \dfrac{{ - \dfrac{\pi }{4}}}{{\dfrac{1}{2}\left( {\dfrac{\pi }{4} + 1} \right)}} + 1 = \dfrac{{ - 2\pi }}{{\left( {\pi  + 4} \right)}} + 1 = \dfrac{{4 - \pi }}{{4 + \pi }} \Rightarrow m = 4\end{array}$. 

Ta có: \(x.f\left( {{x^3}} \right) + f\left( {{x^2} - 1} \right) = {e^{{x^2}}}\) \( \Leftrightarrow {x^2}.f\left( {{x^3}} \right) + xf\left( {{x^2} - 1} \right) = x{e^{{x^2}}}\).

Lấy tích phân tư -1 đến 0 hai vế phương trình ta có:

\(\int\limits_{ - 1}^0 {{x^2}.f\left( {{x^3}} \right)dx}  + \int\limits_{ - 1}^0 {xf\left( {{x^2} - 1} \right)dx}  = \int\limits_{ - 1}^0 {x{e^{{x^2}}}dx} \,\,\left( * \right)\).

Xét \({I_1} = \int\limits_{ - 1}^0 {{x^2}.f\left( {{x^3}} \right)dx} \).

Đặt \(t = {x^3} \Rightarrow dt = 3{x^2}dx \Rightarrow {x^2}dx = \dfrac{{dt}}{3}\).

Đổi cận: \(\left\{ \begin{array}{l}x =  - 1 \Rightarrow t =  - 1\\x = 0 \Rightarrow t = 0\end{array} \right.\), khi đó ta có: \({I_1} = \dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( t \right)dt}  = \dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( x \right)dx} \).

Xét \({I_2} = \int\limits_{ - 1}^0 {xf\left( {{x^2} - 1} \right)dx} \).

Đặt \(u = {x^2} - 1 \Rightarrow du = 2xdx \Rightarrow xdx = \dfrac{1}{2}du\).

Đổi cận: \(\left\{ \begin{array}{l}x =  - 1 \Rightarrow u = 0\\x = 0 \Rightarrow u =  - 1\end{array} \right.\), khi đó ta có \({I_2} = \dfrac{1}{2}\int\limits_0^{ - 1} {f\left( u \right)du}  =  - \dfrac{1}{2}\int\limits_{ - 1}^0 {f\left( x \right)dx} \).

Xét \({I_3} = \int\limits_{ - 1}^0 {x{e^{{x^2}}}dx} \)

Đặt \(v = {x^2} \Rightarrow dv = 2xdx \Rightarrow xdx = \dfrac{1}{2}dv\).

Đổi cận: \(\left\{ \begin{array}{l}x =  - 1 \Rightarrow v = 1\\x = 0 \Rightarrow v = 0\end{array} \right.\), khi đó ta có \({I_3} = \dfrac{1}{2}\int\limits_1^0 {{e^v}dv}  = \dfrac{1}{2}\left. {{e^v}} \right|_1^0 = \dfrac{1}{2} - \dfrac{e}{2} = \dfrac{{1 - e}}{2}\).

Thay tất cả vào (*) ta có:

\(\begin{array}{l}\dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( x \right)dx}  - \dfrac{1}{2}\int\limits_{ - 1}^0 {f\left( x \right)dx}  = \dfrac{{1 - e}}{2}\\ \Leftrightarrow  - \dfrac{1}{6}\int\limits_{ - 1}^0 {f\left( x \right)dx}  = \dfrac{{1 - e}}{2}\\ \Leftrightarrow \int\limits_{ - 1}^0 {f\left( x \right)dx}  = 3\left( {e - 1} \right)\end{array}\)

Bước 1: Sử dụng tích phân từng phần.

Ta có \(A = \int_0^2 {xf'\left( x \right)dx} \)

Đặt \(\left\{ \begin{array}{l}x = u\\dv = f'\left( x \right)dx\end{array} \right. \)\(\Rightarrow \left\{ \begin{array}{l}dx = du\\v = f\left( x \right)\end{array} \right.\).

Khi đó \(A = \left. {x.f\left( x \right)} \right|_0^2 - \int\limits_0^2 {f\left( x \right)dx} \)\( = 2f\left( 2 \right) - \int\limits_0^2 {f\left( x \right)dx} \).

Bước 2: Sử dụng phương pháp đổi biến số.

Xét \(B = \int\limits_0^1 {f\left( {2x} \right)dx} \).

Đặt \(t = 2x \Rightarrow dt = 2dx\).

Đổi cận \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 2\end{array} \right.\)

Khi đó ta có \(B = \dfrac{1}{2}\int\limits_0^2 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^2 {f\left( x \right)dx} \)\( = 2 \)\(\Rightarrow \int\limits_0^2 {f\left( x \right)dx}  = 4\)

Vậy \(A = 2.1 - 4 =  - 2\).