Nguyên hàm (phương pháp đổi biến)

Ta có: \(I = \int\limits_0^4 {\sin \sqrt x dx} \)

Đặt \(u = \sqrt x  \Rightarrow {u^2} = x \Rightarrow dx = 2udu\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow u = 0\\x = 4 \Rightarrow u = 2\end{array} \right..\)

\( \Rightarrow I = \int\limits_0^2 {2u\sin udu} .\)

Đặt \(I = \int {\dfrac{{{e^x}}}{{\sqrt {{e^x} + 1} }}dx} \)

Đặt \(t = \sqrt {{e^x} + 1}  \Rightarrow {t^2} = {e^x} + 1\) \( \Rightarrow 2tdt = {e^x}dx\).

Khi đó ta có: \(I = \int {\dfrac{{2tdt}}{t} = \int {2dt.} } \)

Đặt \(t = 2x - 3 \Rightarrow dt = 2xdx\).

Khi đó ta có: \(\int {f\left( {2x - 3} \right)dx}  = \dfrac{1}{2}\int {f\left( t \right)dt} \).

Mà \(\int {f\left( x \right)dx}  = F\left( x \right) + C\) nên \(\int {f\left( t \right)dt}  = F\left( t \right) + C = F\left( {2x - 3} \right) + C\).

Vậy \(\int {f\left( {2x - 3} \right)dx}  = \dfrac{1}{2}F\left( {2x - 3} \right) + C\).

Ta có: \(u\left( t \right) = v\left( x \right) \Rightarrow u'\left( t \right)dt = v'\left( x \right)dx \Rightarrow dt = \dfrac{{v'\left( x \right)}}{{u'\left( t \right)}}dx\)

Đặt $t = \sqrt {3{x^2} + 1}  \Rightarrow 2tdt = 6xdx \Rightarrow \dfrac{1}{3}tdt = xdx$

$I = \dfrac{1}{3}\int {{t^2}} dt = \dfrac{1}{9}{t^3} + C = \dfrac{1}{9}\sqrt {{{\left( {3{x^2} + 1} \right)}^3}}  + C$

Vậy $a = 9;b = 3$

Ta có:  \(t = {x^2}\)\( \Rightarrow dt = 2xdx\) hay \(\int {2xf\left( {{x^2}} \right)dx}  = \int {f\left( t \right)dt} \)

Đặt \(t = \sqrt {{x^2} + 1}  \Leftrightarrow {x^2} = {t^2} - 1 \Rightarrow xdx = tdt\)

\(A = \int {{{\left( {{t^2} - 1} \right)}^2}{t^2}dt}  = \int {\left( {{t^6} - 2{t^4} + {t^2}} \right)dt} \)\( = \dfrac{{{t^7}}}{7} - \dfrac{2}{5}{t^5} + \dfrac{{{t^3}}}{3} + C\) \( \Rightarrow a = \dfrac{1}{7};b =  - \dfrac{2}{5};c = \dfrac{1}{3}\) \( \Rightarrow a - b - c = \dfrac{{22}}{{105}}\)

$f(x) = {x^2}{e^{{x^3} + 1}}$ $ \Rightarrow \int {f(x)dx = \int {{x^2}} {e^{{x^3} + 1}}dx = \dfrac{1}{3}\int {{e^{{x^3} + 1}}} d({x^3} + 1) } $ $= \dfrac{1}{3}{e^{{x^3} + 1}} + C.$

Đặt $t = \sqrt {{e^x} - 1}  \Rightarrow {t^2} = {e^x} - 1 \Rightarrow \left\{ \begin{array}{l}2tdt = {e^x}dx\\{e^x} = {t^2} + 1\end{array} \right.$

            $ \Rightarrow I = \int\limits_{}^{} {\dfrac{{{e^x}.{e^x}dx}}{{\sqrt {{e^x} - 1} }}}  = \int\limits_{}^{} {\dfrac{{\left( {{t^2} + 1} \right).2tdt}}{t}} $ $ = 2\int\limits_{}^{} {\left( {{t^2} + 1} \right)dt}  = 2\left( {\dfrac{{{t^3}}}{3} + t} \right) + C$ $ \Rightarrow a = \dfrac{2}{3};b = 2 \Rightarrow {a^2} + {b^2} = \dfrac{{40}}{9}$

Ta có $F\left( x \right) = \int {f\left( x \right){\mkern 1mu} {\rm{d}}x}  = \int {\dfrac{{2017x}}{{{{\left( {{x^2} + 1} \right)}^{2018}}}}{\rm{d}}x}  = \dfrac{{2017}}{2}\int {\dfrac{{{\rm{d}}\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^{2018}}}}}  =  - \dfrac{1}{{2{{\left( {{x^2} + 1} \right)}^{2017}}}} + C.$

Mà $F\left( 1 \right) = 0$$ \Rightarrow $$C - \dfrac{1}{{{2^{2018}}}} = 0 \Leftrightarrow C = \dfrac{1}{{{2^{2018}}}}.$

Khi đó $F\left( x \right) =  - \dfrac{1}{{2{{\left( {{x^2} + 1} \right)}^{2017}}}} + \dfrac{1}{{{2^{2018}}}}.$

Mặt khác ${\left( {{x^2} + 1} \right)^{2017}} \ge 1 \Leftrightarrow  - \dfrac{1}{{2{{\left( {{x^2} + 1} \right)}^{2017}}}} \ge  - {\mkern 1mu} \dfrac{1}{2}$ suy ra $F\left( x \right) \ge  - {\mkern 1mu} \dfrac{1}{2} + \dfrac{1}{{{2^{2018}}}} \Rightarrow m = \dfrac{{1 - {2^{2017}}}}{{{2^{2018}}}}.$

Đặt \(t = \sqrt {1 + 3\cos x} \) \( \Rightarrow {t^2} - 1 = 3\cos x \Rightarrow 2tdt =  - 3\sin xdx\)

Lại có: \(\sin 2x + \sin x = 2\sin x\cos x + \sin x = \left( {2\cos x + 1} \right)\sin x\)

Do đó \(\dfrac{{\sin 2x + \sin x}}{{\sqrt {1 + 3\cos x} }}dx = \dfrac{{\left( {2\cos x + 1} \right)\sin xdx}}{{\sqrt {1 + 3\cos x} }}\) \( = \dfrac{{\left( {2.\dfrac{{{t^2} - 1}}{3} + 1} \right).\dfrac{{ - 2tdt}}{3}}}{t} =  - \dfrac{2}{9}\left( {2{t^2} + 1} \right)dt\)

$ \Rightarrow I =  - \int {\dfrac{2}{9}\left( {2{t^2} + 1} \right)dt}  =  - \dfrac{2}{9}\left( {\dfrac{{2{t^3}}}{3} + t} \right) + C$$ =  - \dfrac{2}{9}\left( {\dfrac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C$

$ \Rightarrow F\left( x \right) =  - \dfrac{2}{9}\left( {\dfrac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C$

$ \Rightarrow F\left( {\dfrac{\pi }{2}} \right) - F\left( 0 \right) = \dfrac{{34}}{{27}}$

Ta có $F\left( x \right) = \int {f\left( x \right){\mkern 1mu} dx}  = \int {\dfrac{{dx}}{{2{e^x} + 3}}}  = \int {\dfrac{{{e^x}}}{{2{e^{2x}} + 3{e^x}}}dx} .$

Đặt ${e^x} = t \Rightarrow {e^x}dx = dt.$

$\begin{array}{*{20}{l}}{\rm{\;}}&{ \Rightarrow I = \int {\dfrac{{dt}}{{2{t^2} + 3t}} = \dfrac{1}{3}\int {\left( {\dfrac{1}{t} - \dfrac{2}{{2t + 3}}} \right)dt = \dfrac{1}{3}\left( {\ln t - \ln \left( {2t + 3} \right)} \right)} }  + C}\\{{\rm{ \;}}}&{ \Rightarrow F\left( x \right) = \dfrac{1}{3}\left( {x - \ln \left( {2{e^x} + 3} \right)} \right) + C.}\end{array}$

Mà $F\left( 0 \right) = 10$ suy ra $C - \dfrac{{\ln 5}}{3} = 10 \Leftrightarrow C = 10 + \dfrac{{\ln 5}}{3}.$

Vậy $F\left( x \right) = \dfrac{1}{3}\left( {x - \ln \left( {2{e^x} + 3} \right)} \right) + 10 + \dfrac{{\ln 5}}{3}.$

Ta có: \(\int {\sqrt[6]{{1 - {{\cos }^3}x}}\sin x{{\cos }^5}xdx}  = \int {\sqrt[6]{{1 - {{\cos }^3}x}}.{{\cos }^3}x.\sin x.{{\cos }^2}xdx} \)

Đặt \(t = \sqrt[6]{{1 - {{\cos }^3}x}}\) \( \Rightarrow {t^6} = 1 - {\cos ^3}x\) \( \Rightarrow 6{t^5}dt = 3\sin x{\cos ^2}xdx\) và \({\cos ^3}x = 1 - {t^6}\).

$\int {t.(1 - {t^6}).2{t^5}.dt}  = \int {(2{t^6} - 2{t^{12}}).dt}  = \dfrac{{2{t^7}}}{7} - {\dfrac{{2t}}{{13}}^{13}} + C = 2\left( {\dfrac{{{t^7}}}{7} - \dfrac{{{t^{13}}}}{{13}}} \right) + C$

\( \Rightarrow \alpha  = 7;\beta  = 13 \Rightarrow \dfrac{\alpha }{\beta } = \dfrac{7}{{13}}\)

$\int {{x^2}\sqrt {4 + {x^3}} dx = \dfrac{1}{3}\int {\sqrt {4 + {x^3}} .d\left( {{x^3} + 4} \right)} } $$ = \dfrac{1}{3}\dfrac{{{{\left( {4 + {x^3}} \right)}^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + C = \dfrac{2}{9}\sqrt {{{\left( {4 + {x^3}} \right)}^3}}  + C$.

Đặt \(t = \sqrt {{x^2} + 3}  \Rightarrow {t^2} = {x^2} + 3 \Rightarrow 2tdt = 2xdx \Rightarrow xdx = tdt\).

Suy ra $I = \int {t.tdt = \int {{t^2}dt = \dfrac{{{t^3}}}{3}} }  + C$$ = \dfrac{{{{\left( {\sqrt {{x^2} + 3} } \right)}^3}}}{3} + C = \dfrac{{\sqrt {{{\left( {{x^2} + 3} \right)}^3}} }}{3} + C$

 Vậy  \(S = \log _3^23 + {\log _3}3 + 2016 = 2018\)

Đặt \(t = \sqrt {2{\rm{x}} - 1}  \Rightarrow {t^2} = 2{\rm{x}} - 1 \Rightarrow t{\rm{d}}t = d{\rm{x}}\)\( \Rightarrow I = \int {\dfrac{{t{\rm{d}}t}}{{t + 4}} = \int {\left( {1 - \dfrac{4}{{t + 4}}} \right)dt = t - 4\ln \left| {t + 4} \right| + C} } \)

$ = \sqrt {2{\rm{x}} - 1}  - \ln {\left( {\sqrt {2{\rm{x}} - 1}  + 4} \right)^4} + C$

Vậy $n = 4\;$ suy ra \(S = \sin \dfrac{{4\pi }}{8} = 1\)

Đặt \(t = \sqrt {\ln x + 1} \) \( \Rightarrow {t^2} = \ln x + 1 \Rightarrow 2tdt = \dfrac{1}{x}dx\) và \(\ln x = {t^2} - 1\)

\(I = \int {\dfrac{{{{\left( {{t^2} - 1} \right)}^2}}}{t}.2tdt} \)\( = \int {2\left( {{t^4} - 2{t^2} + 1} \right)dt} \) \( = 2\left( {\dfrac{{{t^5}}}{5} - \dfrac{{2{t^3}}}{3} + t} \right) + C\) \( = \dfrac{2}{{15}}\left( {3{t^5} - 10{t^3} + 15t} \right) + C\)  \( \Rightarrow \left\{ \begin{array}{l}b = 3\\c =  - 10\\d = 15\end{array} \right.\)

Vậy \(A = \dfrac{2}{{15}}.3.\left( { - 10} \right).15 =  - 60\)

Theo công thức đổi biến số ta được \(f\left( x \right)dx = f\left( {u\left( t \right)} \right).u'\left( t \right)dt\)

\(\int {\dfrac{{\sin 2xdx}}{{\cos 2x - 1}}}  = \int {\dfrac{{2\sin x\cos x}}{{1 - 2{{\sin }^2}x - 1}}dx}  =  - \int {\dfrac{{\cos x}}{{\sin x}}dx}  =  - \int {\dfrac{{d\left( {\sin x} \right)}}{{\sin x}}}  =  - \ln \left| {\sin x} \right| + C\)

Ta có: \(x = \sin t \Rightarrow dx = \left( {\sin t} \right)'dt = \cos tdt\)