Sử dụng phương pháp đổi biến số để tìm nguyên hàm

Nếu \(t = u\left( x \right)\)thì \(dt = u'\left( x \right)dx\).

Đặt $t = 3x \Rightarrow dt = 3dx \Rightarrow dx = \dfrac{{dt}}{3}$, khi đó:

$\begin{array}{*{20}{l}}{\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x} {\rm{\;}} = \dfrac{1}{3}\int {f\left( t \right)dt} {\rm{\;}} = \dfrac{1}{3}\left( {2t\ln \left( {3t - 1} \right)} \right) + C}\\{ = \dfrac{1}{3}\left( {2.3x.\ln \left( {3.3x - 1} \right)} \right) + C = 2x\ln \left( {9x - 1} \right) + C}\end{array}$

Vậy $\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x} {\rm{\;}} = 2x\ln \left( {9x - 1} \right) + C$

Ta có: \(t = {x^2} \Rightarrow dt = 2xdx \Rightarrow xdx = \dfrac{{dt}}{2} \)

$\Rightarrow xf\left( {{x^2}} \right)dx = f\left( {{x^2}} \right).xdx = f\left( t \right).\dfrac{{dt}}{2} = \dfrac{1}{2}f\left( t \right)dt$

Ta có: \(\sqrt {1 - {{\cos }^2}x}  = t \)

$\Rightarrow {t^2} = 1 - {\cos ^2}x \Rightarrow 2tdt = 2\cos x\sin xdx = \sin 2xdx \Rightarrow \sin 2xdx = 2tdt$

Suy ra \(f\left( x \right)dx = \sin 2x\sqrt {1 - {{\cos }^2}x} dx = \sqrt {1 - {{\cos }^2}x} .\sin 2xdx = t.2tdt = 2{t^2}dt\)

\(I = \int {3{x^5}\sqrt {{x^3} + 1} dx}  = \int {3{x^2}.{x^3}} \sqrt {{x^3} + 1} dx\)

Đặt \(\sqrt {{x^3} + 1}  = t \Rightarrow {x^3} + 1 = {t^2} \Rightarrow 3{x^2}dx = 2tdt\)

\( \Rightarrow I = \int {\left( {{t^2} - 1} \right).t.2tdt = 2\int {\left( {{t^4} - {t^2}} \right)dt = \dfrac{2}{5}{t^5} - \dfrac{2}{3}{t^3} + C} }  \)

$= \dfrac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1}  - \dfrac{2}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1}  + C$

\(F\left( x \right) = \int {\dfrac{{\ln x}}{{x\sqrt {1 - \ln x} }}dx} \)

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

\( \Rightarrow F\left( x \right) = \int {\dfrac{{1 - {t^2}}}{t}\left( { - 2tdt} \right) =  - 2\int {\left( {1 - {t^2}} \right)} } dt \)

$=  - 2t + \dfrac{2}{3}{t^3} + C =  - 2\sqrt {1 - \ln x}  + \dfrac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + C$

\(\begin{array}{l}F\left( e \right) =  - 2\sqrt {1 - 1}  +\dfrac{2}{3}\left( {1 - 1} \right)\sqrt {1 - 1}  + C = 3 \Rightarrow C = 3\\ \Rightarrow F\left( x \right) =  - 2\sqrt {1 - \ln x}  + \dfrac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + 3\end{array}\)

\(I = \int {\dfrac{{{{\cos }^3}x}}{{1 + \sin x}}dx}  = \int {\dfrac{{{{\cos }^2}x.\cos xdx}}{{1 + \sin x}} = \int {\dfrac{{\left( {1 - {{\sin }^2}x} \right)\cos xdx}}{{1 + \sin x}}} } \) 

Đặt \(\sin x = t \Rightarrow \cos xdx = dt\)\(I = \int {\dfrac{{\left( {1 - {t^2}} \right)dt}}{{1 + t}} = \int {\left( {1 - t} \right)dt = t - \dfrac{1}{2}{t^2} + C} } \)

\(f\left( x \right) = \dfrac{{{x^2}}}{{\sqrt {1 - x} }}\) và \(t = \sqrt {1 - x}  \Rightarrow 1 - x = {t^2} \Rightarrow x = 1 - {t^2} \Rightarrow dx =  - 2tdt\)

\( \Rightarrow \int {f\left( x \right)} dx = \int {\dfrac{{{{\left( {1 - {t^2}} \right)}^2}}}{t}\left( { - 2tdt} \right) =  - 2\int {{{\left( {1 - {t^2}} \right)}^2}dt}  =  - 2\int {{{\left( {{t^2} - 1} \right)}^2}dt} }  \)

$\Rightarrow m = 1$

\(F\left( x \right) = \int {\dfrac{x}{{1 + \sqrt {1 + x} }}dx} \)

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

\(\begin{array}{l} \Rightarrow F\left( x \right) = \int {\dfrac{{{t^2} - 1}}{{1 + t}}.2tdt = 2\int {t\left( {t - 1} \right)dt = 2\int {\left( {{t^2} - t} \right)} } } dt \\ = \dfrac{2}{3}{t^3} - {t^2} + C = \dfrac{2}{3}\left( {1 + x} \right)\sqrt {1 + x}  - \left( {1 + x} \right) + C\\ \Rightarrow F\left( 3 \right) - F\left( 0 \right) = \dfrac{2}{3}\left( {1 + 3} \right)\sqrt {1 + 3}  - \left( {1 + 3} \right) - \dfrac{2}{3}\left( {1 + 0} \right)\sqrt {1 + 0}  + \left( {1 + 0} \right) = \dfrac{5}{3}\\ \Rightarrow a = 5,b = 3 \Rightarrow a + b = 8\end{array}\)

\(I = \int {\dfrac{{6{\mathop{\rm tanx}\nolimits} }}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx} \)

Đặt \(u = \sqrt {3\tan x + 1}  \Rightarrow {u^2} = 3\tan x + 1 \Rightarrow \dfrac{3}{{{{\cos }^2}x}}dx = 2udu \Rightarrow \dfrac{{dx}}{{{{\cos }^2}x}} = \dfrac{{2udu}}{3}\)\(I = \int {\dfrac{{2\left( {{u^2} - 1} \right)}}{{3u}}2udu = \dfrac{4}{3}\int {\left( {{u^2} - 1} \right)} } du\)

\(I = \int {\dfrac{{{e^{2x}}}}{{\left( {{e^x} + 1} \right)\sqrt {{e^x} + 1} }}} dx = a\left( {t + \dfrac{1}{t}} \right) + C\)

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

\(I = \int {\dfrac{{{t^2} - 1}}{{{t^2}.t}}2tdt }= 2\int {\left( {1 - \dfrac{1}{{{t^2}}}} \right)dt}  \) \(= 2\left( {t + \dfrac{1}{t}} \right) + C  \Rightarrow a = 2\)

Nếu \(x = u\left( t \right)\) thì \(dx = u'\left( t \right)dt\).

\(\int {\cot xdx = \int {\dfrac{{\cos x}}{{\sin x}}dx} } \)

Đặt \(t = \sin x \Rightarrow dt = \cos xdx\).

Khi đó ta có:

\(\begin{array}{l}\int {\cot xdx = \int {\dfrac{{\cos x}}{{\sin x}}dx} }  = \int {\dfrac{{dt}}{t}}  = \ln \left| t \right| + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| {\sin x} \right| + C\end{array}\)

\(\int {\sin x.\cos 2xdx}  = \int {\left( {2{{\cos }^2}x - 1} \right)\sin xdx}  =  - \int {\left( {2{{\cos }^2}x - 1} \right)d\left( {\cos x} \right)}  = \dfrac{{ - 2{{\cos }^3}x}}{3} + \cos x + C\)

Ta có: \(x = \cot t \Rightarrow dx = \left( {\cot t} \right)'dt =  - \dfrac{1}{{{{\sin }^2}t}}dt =  - \left( {1 + {{\cot }^2}t} \right)dt\)

Do

\(\begin{array}{l}\dfrac{1}{{{{\sin }^2}t}} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}}\\ = 1 + {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} = 1 + {\cot ^2}x\end{array}\)

Ta có: \(x = \tan t \Rightarrow dx=\dfrac{1}{{{{\cos }^2}t}} dt = \left( {1 + {{\tan }^2}t} \right)dt\).

Do đó \(f\left( x \right)dx = \dfrac{1}{{{x^2} + 1}}dx = \dfrac{1}{{{{\tan }^2}t + 1}}\left( {1 + {{\tan }^2}t} \right)dt = dt\)

Đặt \(t = \sqrt {8 - {x^2}}  \Rightarrow {t^2} = 8 - {x^2} \Rightarrow  - tdt = xdx\)

\(\int {\dfrac{x}{{\sqrt {8 - {x^2}} }}dx =  - \int {\dfrac{{tdt}}{t} =  - t + C =  - \sqrt {8 - {x^2}}  + C} } \)

Vì \(F\left( 2 \right) = 0\) nên \(C = 2\)

Ta có phương trình $ - \sqrt {8 - {x^2}} + 2 = x \Leftrightarrow x = 1 - \sqrt 3 $

Ta có $f\left( x \right) = \sqrt {3 - 2x - {x^2}}  = \sqrt {4 - \left( {1 + 2x + {x^2}} \right)}  = \sqrt {4 - {{\left( {x + 1} \right)}^2}} .$

Đặt $x + 1 = 2\sin t \Leftrightarrow {\rm{d}}x = 2\cos t\,{\rm{d}}t$ và $4 - {\left( {x + 1} \right)^2} = 4 - 4{\sin ^2}t = 4{\cos ^2}t$

Do $0\le t \le \dfrac{\pi }{2}$ nên $\cos t \ge 0$.

Khi đó $\int {f\left( x \right)\,{\rm{d}}x}  = \int {\sqrt {4{{\cos }^2}t} .2\cos t\,\,{\rm{d}}t}  = 4\int {{{\cos }^2}t\,{\rm{d}}t}  = 2\int {\left( {1 + \cos 2t} \right){\rm{d}}t} $.

Đặt \(u = 5x + 2 \Rightarrow du = 5dx\).

$ \Rightarrow \int {f(5x + 2)dx}  = \int {f\left( u \right).\dfrac{1}{5}du}  = \dfrac{1}{5}\int {f\left( u \right)du}  $

$= \dfrac{1}{5}F\left( u \right) + C = \dfrac{1}{5}F\left( {5x + 2} \right) + C$

\(\int {\dfrac{x}{{\sqrt {3{{\rm{x}}^2} + 2} }}d{\rm{x}}}  = \dfrac{1}{6}\int {\dfrac{{d\left( {3{{\rm{x}}^2} + 2} \right)}}{{\sqrt {3{{\rm{x}}^2} + 2} }}}  \) \(= \dfrac{1}{3}\int {\dfrac{{d\left( {3{{\rm{x}}^2} + 2} \right)}}{{2\sqrt {3{{\rm{x}}^2} + 2} }}}  = \dfrac{1}{3}\sqrt {3{{\rm{x}}^2} + 2}  + C\)