Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = f(x)}\\{dv = g\prime (x)dx}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{du = f\prime (x)dx}\\{v = g(x)}\end{array}} \right.\) khi đó

\[I = f\left( x \right).g\left( x \right)\left| {_a^b} \right. - \int\limits_a^b {f'\left( x \right)} .g\left( x \right)dx\]

Đáp án cần chọn là: C

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = {x^2}}\\{dv = cosxdx}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{du = 2xdx}\\{v = sinx}\end{array}} \right.\)khi đó\[I = {x^2}sinx\left| {_0^{\frac{\pi }{2}}} \right. - 2\int\limits_0^{\frac{\pi }{2}} {xsinxdx} \]

Đáp án cần chọn là: B

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

Khi đó

\[\int\limits_0^1 {g\left( x \right)} .f'\left( x \right)dx = \left[ {g(x).f(x)} \right]\left| {_0^1} \right. - \int\limits_0^1 {g'\left( x \right)} .f\left( x \right)dx\]

\( \Leftrightarrow \left[ {g(x).f(x)} \right]\left| {_0^1} \right. = 3\)

Mặt khác\(I = \int\limits_0^1 {\left[ {f\left( x \right).g\left( x \right)} \right]} 'dx = \left[ {f\left( x \right).g\left( x \right)} \right]\left| {_0^1} \right. \Rightarrow I = 3\)

Đáp án cần chọn là: C

Đáp án cần chọn là: A

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

Khi đó\[I = - \frac{{x + lnx}}{{2{{(x + 1)}^2}}}\left| {_1^2} \right. + \int\limits_1^2 {\frac{{x + 1}}{x}.\frac{1}{{2{{(x + 1)}^2}}}} dx\]

\[ = - \frac{{2 + \ln 2}}{{18}} + \frac{1}{8} + \frac{1}{2}\mathop \smallint \limits_1^2 \frac{{{\rm{d}}x}}{{x\left( {x + 1} \right)}}\]

\[ = - \frac{{2 + \ln 2}}{{18}} + \frac{1}{8} + \frac{1}{2}\mathop \smallint \limits_1^2 \left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right){\rm{d}}x.\]

\( = - \frac{{2 + ln2}}{{18}} + \frac{1}{8} + \frac{1}{2}(ln|x| - ln|x + 1|)\left| {_1^2} \right.\)

\(\begin{array}{l} = \frac{1}{{72}} - \frac{1}{{18}}ln2 + \frac{1}{2}(ln2 - ln3 + ln2)\\ = \frac{1}{{72}} + \frac{{17}}{{18}}ln2 - \frac{1}{2}\ln 3\\ = a + b.ln2 - c.ln3\end{array}\)

Vậy\(\left\{ {\begin{array}{*{20}{c}}{a = \frac{1}{{72}}}\\{b = \frac{{17}}{{18}}}\\{c = \frac{1}{2}}\end{array}} \right. \Rightarrow \frac{c}{a} = \frac{1}{2}:\frac{1}{{72}} = 36\)

Đáp án cần chọn là: D

Ta có :\[{\left( {x\sin x + \cos x} \right)^\prime } = \sin x + x\cos x - \sin x = x\cos x\]

\[ \Rightarrow I = \mathop \smallint \limits_0^{\frac{\pi }{4}} \frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}{\rm{d}}x = \mathop \smallint \limits_0^{\frac{\pi }{4}} \frac{{\frac{x}{{\cos x}}.x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dv\]

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

Khi đó

\(I = - \frac{x}{{cosx}}.\frac{1}{{xsinx + cosx}}\left| {_0^{\frac{\pi }{4}}} \right. + \int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{co{s^2}x}}} \)

\( = \frac{{ - \frac{\pi }{4}}}{{\frac{{\sqrt 2 }}{2}}}.\frac{1}{{\frac{\pi }{4}\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}}} + tanx\left| {_0^{\frac{\pi }{4}}} \right.\)

\( = \frac{{ - \frac{\pi }{4}}}{{\frac{1}{2}\left( {\frac{\pi }{4} + 1} \right)}} + 1 = \frac{{ - 2\pi }}{{\left( {\pi + 4} \right)}} + 1 = \frac{{4 - \pi }}{{4 + \pi }} \Rightarrow m = 4\)

Đáp án cần chọn là: C

Đặt

\(\left\{ {\begin{array}{*{20}{c}}{u = ln(3sinx + cosx)}\\{dv = \frac{{dx}}{{si{n^2}x}}}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{{3cosx - sinx}}{{3sinx + cosx}}dx}\\{v = - cotx - 3 = - \frac{{3sinx + cosx}}{{sinx}}}\end{array}} \right.\)

Khi đó

\[I = [ - (cotx + 3)ln(3sinx + cosx)]\left| {_{\frac{\pi }{4}}^{\frac{\pi }{2}}} \right. + \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{3cosx - sinx}}{{sinx}}} dx\]

\[ = 4.ln2\sqrt 2 - 3.ln3 - \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {dx + 3\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{d(sinx)}}{{sinx}}} .} \]

\[ = 4.ln2\sqrt 2 - 3.ln3 - \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {dx + 3ln|sinx|\left| {_{\frac{\pi }{4}}^{\frac{\pi }{2}}} \right.} \]

\[\begin{array}{l} = 4.ln2\sqrt 2 - 3.ln3 - \frac{\pi }{4} - 3.ln\frac{1}{{\sqrt 2 }}\\ = 12ln\sqrt 2 - 3ln3 - \frac{\pi }{4} + 3ln\sqrt 2 = 15.ln\sqrt 2 - 3.ln3 - \frac{\pi }{4}\end{array}\]

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{m = 15}\\{n = - 3}\end{array}} \right. \Rightarrow m + n = 12\)

Đáp án cần chọn là: A

Đáp án cần chọn là: D

Dùng máy tính kiểm tra từng đáp án hoặc:

Đặt\[u = \ln x,dv = xdx \Rightarrow du = \frac{{dx}}{x},v = \frac{{{x^2}}}{2}\]

\(I = \frac{{{x^2}lnx}}{2}\left| {_1^e} \right. - \int\limits_1^e {\frac{x}{2}} dx = \frac{{{e^2}}}{2} - \frac{{{x^2}}}{4}\left| {_1^e} \right. = \frac{{{e^2}}}{2} - \left( {\frac{{{e^2}}}{2} - \frac{1}{4}} \right) = \frac{{{e^2} + 1}}{4}\)

Đáp án cần chọn là: C

Đáp án cần chọn là: A

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(x) là một nguyên hàm của hàm số \[{e^{2x}}\cos x\], điều kiện là\[f\prime (x) = {e^{2x}}cos3x \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{2a + 3b = 1}\\{2b - 3a = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = \frac{2}{{13}}}\\{b = \frac{3}{{13}}}\end{array} \Rightarrow a + b = \frac{5}{{13}}} \right.\]

Đáp án cần chọn là: C

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

Ta có:

\[\int\limits_0^1 {(x + 1)f\prime (x)dx = 10 \Leftrightarrow (x + 1)f(x)\left| {_0^1} \right.} - \int\limits_0^1 {f(x)dx = 10 = 2f(1) - f(0) - } \int\limits_0^1 {f(x)dx} \]

\( \Rightarrow \int\limits_0^1 {f(x)dx = - 8.} \)

Đáp án cần chọn là: D

Đáp án cần chọn là: B

Đáp án cần chọn là: B

Đáp án cần chọn là: A

Đáp án cần chọn là: B

Đáp án cần chọn là: A

Đáp án cần chọn là: A

Đáp án cần chọn là: A

Đáp án cần chọn là: B

Đáp án cần chọn là: A

Xét tích phân\[I = \mathop \smallint \limits_0^{{\pi ^2}} f\left( {\sqrt x } \right)\cos \left( {\sqrt x } \right)dx\]

Đặt\[t = \sqrt x \Rightarrow {t^2} = x \Rightarrow 2tdt = dx\]

Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = 0 \Rightarrow t = 0}\\{x = {\pi ^2} \Rightarrow t = \pi }\end{array}} \right.\) khi đó ta có

\[I = \mathop \smallint \limits_0^\pi f\left( t \right)\cos \left( t \right)2tdt = \mathop \smallint \limits_0^\pi 2f\left( x \right)\cos x.xdx\]

Xét tích phân\[\mathop \smallint \limits_0^\pi xf'\left( x \right)\sin xdx = 5\]

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

\[\begin{array}{l}\mathop \smallint \limits_0^\pi xf\left( x \right)'\sin xdx = 5\\ \Leftrightarrow (xsinx.f(x))\left| {_0^\pi } \right. - \int\limits_0^\pi {[f(x)sinx + xf(x)cosx]dx = 5} \end{array}\]

\( \Leftrightarrow - \int\limits_0^\pi {f(x)sinxdx - \int\limits_0^\pi {xf(x)cosxdx = 5} } \)

\[\begin{array}{l} \Leftrightarrow - 20 - \frac{I}{2} = 5\\ \Leftrightarrow \frac{I}{2} = - 25\\ \Leftrightarrow I = - 50\end{array}\]

Đáp án cần chọn là: B

Ta có:\[\mathop \smallint \limits_1^3 \left( {4 - x} \right)f\left( x \right)dx = 4\mathop \smallint \limits_1^3 f\left( x \right)dx - \mathop \smallint \limits_1^3 xf\left( x \right)dx\]

Đặt\[t = 4 - x \Rightarrow dt = - dx\]

Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = 1 \Rightarrow t = 3}\\{x = 3 \Rightarrow t = 1}\end{array}} \right.\) khi đó ta có:

\[\mathop \smallint \limits_1^3 \left( {4 - x} \right)f\left( x \right)dx = - \mathop \smallint \limits_3^1 tf\left( {4 - t} \right)dt = \mathop \smallint \limits_1^3 tf\left( {4 - t} \right)dt = \mathop \smallint \limits_1^3 tf\left( t \right)dt = \mathop \smallint \limits_1^3 xf\left( x \right)dx\]

\[\begin{array}{*{20}{l}}{ \Rightarrow \mathop \smallint \limits_1^3 xf\left( x \right)dx = 4\mathop \smallint \limits_1^3 f\left( x \right)dx - \mathop \smallint \limits_1^3 xf\left( x \right)dx}\\{ \Leftrightarrow 2\mathop \smallint \limits_1^3 f\left( x \right)dx = \mathop \smallint \limits_1^3 xf\left( x \right)dx = - 2}\end{array}\]

Đáp án cần chọn là: C

Xét tích phân\[\mathop \smallint \limits_4^7 f\left( {\frac{x}{2}} \right)dx = \frac{a}{b}\]

Đặt\[t = \frac{x}{2} \Rightarrow dt = \frac{1}{2}dx \Leftrightarrow dx = 2dt\]  Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = 4 \Rightarrow t = 2}\\{x = 7 \Rightarrow t = \frac{7}{2}}\end{array}} \right.\)

Khi đó ta có:\[I = 2\mathop \smallint \limits_2^{\frac{7}{2}} f\left( t \right)dt\]

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = f(t)}\\{dv = dt}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = f\prime (t)dt}\\{v = t - \frac{7}{2}}\end{array}} \right.\) khi đó ta có:

\[I = 2\left( {\left( {t - \frac{7}{2}} \right)f(t)\mid _2^{\frac{\pi }{2}} - \int\limits_2^{\frac{7}{2}} {\left( {t - \frac{7}{2}} \right)} f\prime (t)dt} \right)\]

\(I = 2\left( {\frac{7}{2}f\left( 0 \right) - \int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right)f'\left( x \right)dx} } \right)\)

\(I = 2\left( {\frac{7}{2}f\left( 2 \right) - \int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right).\frac{{x + 7}}{{\sqrt {2x - 3} }}dx} } \right)\)

\(I = - 2\int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right)} .\frac{{x + 7}}{{\sqrt {2x - 3} }}dx\)

\(I = \frac{{236}}{{15}}\)

\[ \Rightarrow a = 236,b = 15\]

Vậy\[a + b = 236 + 15 = 251\]

Đáp án cần chọn là: B

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

Khi đó ta có:

\(\int\limits_0^{\frac{\pi }{2}} {cosx.f(x)dx = sinx.f(x)\left| {_0^{\frac{\pi }{2}}} \right.} - \int\limits_0^{\frac{\pi }{2}} {xsi{n^2}xdx} \)

\[ = sin\frac{\pi }{2}.f\left( {\frac{\pi }{2}} \right) - \int\limits_0^{\frac{\pi }{2}} {x\frac{{1 - cos2x}}{2}} dx\]

\[ = 2 - \frac{1}{2}\left( {\int\limits_0^{\frac{\pi }{2}} {xdx - \int\limits_0^{\frac{\pi }{2}} {xcos2xdx} } } \right)\]

\[\begin{array}{l} = 2 - \frac{1}{2}\left( {\frac{{{x^2}}}{2}\left| {_0^{\frac{\pi }{2}} - I} \right.} \right)\\ = 2 - \frac{1}{2}\left( {\frac{{{\pi ^2}}}{8} - I} \right)\\ = 2 - \frac{{{\pi ^2}}}{{16}} + \frac{I}{2}\end{array}\]

Xét tích phân\[I = \mathop \smallint \limits_0^{\frac{\pi }{2}} x\cos 2xdx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = cos2xdx}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = \frac{{sin2x}}{2}}\end{array}} \right.\) khi đó ta có:

\[I = x.\frac{{sin2x}}{2}\left| {_0^{\frac{\pi }{2}}} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {sin2xdx} \]

\[I = \frac{\pi }{2}.\frac{{sin\pi }}{2} - 0 + \frac{1}{2}.\frac{{cos2x}}{2}\left| {_0^{\frac{\pi }{2}}} \right.\]

\[I = \frac{1}{4}(cos\pi - cos0)\]

\[I = \frac{1}{4}( - 1 - 1) = - \frac{1}{2}\]

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

\[ \Rightarrow a = 7,\,\,b = 4,\,\,c = 16\]

Vậy\[a + b + c = 7 + 4 + 16 = 27\]Đáp án cần chọn là: D

Xét tích phân: \[I = \mathop \smallint \limits_0^1 \frac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}dx\]

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

Khi đó ta có:

\(\begin{array}{l}I = \frac{{x - 1}}{{2(x + 1)}}f\left( x \right)\left| {_0^1} \right. - \int\limits_0^1 {\frac{{x - 1}}{{2(x + 1)}}f'\left( x \right)} dx\\ \Leftrightarrow I = \frac{1}{2}f\left( 0 \right) - \frac{1}{2}\int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx\\ \Leftrightarrow 4ln2 - 2 = \frac{1}{2}.2 - \frac{1}{2}\int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx\\ \Leftrightarrow \int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx = 6 - 8ln2\end{array}\)

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

\( \Leftrightarrow \int\limits_0^1 {{{[f\prime (x)]}^2}dx + 2k} \int\limits_0^1 {\frac{{x - 1}}{{x + 1}}} f'\left( x \right)dx + {k^2}\int\limits_0^1 {{{\left( {\frac{{x - 1}}{{x + 1}}} \right)}^2}} dx\)

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\int\limits_0^1 {{{\left( {1 - \frac{2}{{x + 1}}} \right)}^2}} dx = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\int\limits_0^1 {\left( {1 - \frac{4}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\left( {x - 4ln|x + 1| - \frac{4}{{x + 1}}} \right)\left| {_0^1} \right. = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}(1 - 4ln2 - 2 + 4) = 0\]

\[ \Leftrightarrow (3 - 4ln2){k^2} - 4(3 - 4ln2)k + 4(3 - 4ln2) = 0\]

\[ \Leftrightarrow {k^2} - 4k + 4 = 0 \Leftrightarrow {(k - 2)^2} = 0 \Leftrightarrow k = 2\]

Khi đó ta có\[\mathop \smallint \limits_0^1 {\left( {f'\left( x \right) - 2.\frac{{x - 1}}{{x + 1}}} \right)^2}dx = 0 \Leftrightarrow f'\left( x \right) = 2.\frac{{x - 1}}{{x + 1}}\]

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

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

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

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

\( \Rightarrow \int\limits_0^1 {f(x)dx = \int\limits_0^1 {[2x - 4ln|x + 1| + 2]dx} } \)

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

Ta có:\[J = \mathop \smallint \limits_0^1 \ln \left| {x + 1} \right|dx = \mathop \smallint \limits_0^1 \ln \left( {x + 1} \right)dx\]

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

\[ \Rightarrow J = (x + 1)ln(x + 1)\left| {_0^1} \right. - \int\limits_0^1 {dx} \]

\[ \Rightarrow J = 2ln2 - 1.ln1 - x\left| {_0^1} \right.\]

\[ \Rightarrow J = 2ln2 - 1\]

Vậy\[\mathop \smallint \limits_0^1 f\left( x \right)dx = 3 - 4\left( {2\ln 2 - 1} \right) = 7 - 8\ln 2\]

Đáp án cần chọn là: D

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:

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

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

Đặt\[t = {x^3} \Rightarrow dt = 3{x^2}dx \Rightarrow {x^2}dx = \frac{{dt}}{3}\]

Đổi cận: \(\left\{ {\begin{array}{*{20}{c}}{x = - 1 \Rightarrow t = - 1}\\{x = 0 \Rightarrow t = 0}\end{array}} \right.\) khi đó ta có:\[{I_1} = \frac{1}{3}\mathop \smallint \limits_{ - 1}^0 f\left( t \right)dt = \frac{1}{3}\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx\]

Xét \[{I_2} = \mathop \smallint \limits_{ - 1}^0 xf\left( {{x^2} - 1} \right)dx\]

Đặt\[u = {x^2} - 1 \Rightarrow du = 2xdx \Rightarrow xdx = \frac{1}{2}du\]

Đổi cận: \(\left\{ {\begin{array}{*{20}{c}}{x = - 1 \Rightarrow u = 0}\\{x = 0 \Rightarrow u = - 1}\end{array}} \right.\) khi đó ta có\[{I_2} = \frac{1}{2}\mathop \smallint \limits_0^{ - 1} f\left( u \right)du = - \frac{1}{2}\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx\]

Xét \[{I_3} = \mathop \smallint \limits_{ - 1}^0 x{e^{{x^2}}}dx\]

Đặt\[v = {x^2} \Rightarrow dv = 2xdx \Rightarrow xdx = \frac{1}{2}dv\]

Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = - 1 \Rightarrow v = 1}\\{x = 0 \Rightarrow v = 0}\end{array}} \right.\) khi đó ta có

\({I_3} = \frac{1}{2}\int\limits_0^1 {{e^v}dv = } \frac{1}{2}{e^v}\left| {_0^1} \right. = \frac{1}{2} - \frac{e}{2} = \frac{{1 - e}}{2}\)

Thay tất cả vào (*) ta có:

\[\begin{array}{*{20}{l}}{\frac{1}{3}\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx - \frac{1}{2}\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = \frac{{1 - e}}{2}}\\{ \Leftrightarrow - \frac{1}{6}\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = \frac{{1 - e}}{2}}\\{ \Leftrightarrow \mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = 3\left( {e - 1} \right)}\end{array}\]

Đáp án cần chọn là: D

Bước 1: Sử dụng tích phân từng phần.

Ta có \[A = \mathop \smallint \nolimits_0^2 xf'\left( x \right)dx\]

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

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

Bước 2: Sử dụng phương pháp đổi biến số.

Xét\[B = \mathop \smallint \limits_0^1 f\left( {2x} \right)dx\]

Đặt\[t = 2x \Rightarrow dt = 2dx\]

Đổi cận\(\left\{ {\begin{array}{*{20}{c}}{x = 0 \Rightarrow t = 0}\\{x = 1 \Rightarrow t = 2}\end{array}} \right.\)

Khi đó ta có\[B = \frac{1}{2}\mathop \smallint \limits_0^2 f\left( t \right)dt = \frac{1}{2}\mathop \smallint \limits_0^2 f\left( x \right)dx = 2 \Rightarrow \mathop \smallint \limits_0^2 f\left( x \right)dx = 4\]

Vậy \[A = 2.1 - 4 = - 2\]