Nguyên hàm (phương pháp đổi biến)

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

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}  $

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

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

Đặt $x = \dfrac{1}{{\sin t}} \Leftrightarrow {\rm{d}}x = {\left( {\dfrac{1}{{\sin t}}} \right)^\prime }{\rm{d}}t \Leftrightarrow {\rm{d}}x =  - \dfrac{{\cos t}}{{{{\sin }^2}t}}{\rm{d}}t$

Và $\dfrac{{\sqrt {{x^2} - 1} }}{{{x^3}}} = {\sin ^3}t.\sqrt {\dfrac{1}{{{{\sin }^2}t}} - 1}  = {\sin ^3}t.\sqrt {\dfrac{{1 - {{\sin }^2}t}}{{{{\sin }^2}t}}}  = {\sin ^3}t.\dfrac{{\cos t}}{{\sin t}} = {\sin ^2}t.\cos t.$

Khi đó $I = \int {{{\sin }^2}t.\cos t.\left( { - \dfrac{{\cos t}}{{{{\sin }^2}t}}} \right){\rm{d}}t}  =  - \,\int {{{\cos }^2}t\,{\rm{d}}t}  =  - \dfrac{1}{2}\int {\left( {1 + \cos 2t} \right){\rm{d}}t} .$

Ta có $f\left( x \right) = \dfrac{{{x^2}\sin x + x\cos x + x\cos x}}{{x\sin x + \cos x}} = x + \dfrac{{x\cos x}}{{x\sin x + \cos x}}$

Khi đó $\int {f\left( x \right){\rm{d}}x}  = \int {\left( {x + \dfrac{{x\cos x}}{{x\sin x + \cos x}}} \right){\rm{d}}x}  = \int {x{\rm{d}}x}  + \int {\dfrac{{x\cos x}}{{x\sin x + \cos x}}{\rm{d}}x.} $

Đặt $t = x\sin x + \cos x \Leftrightarrow {\rm{d}}t = \left( {x\sin x + \cos x} \right){\rm{'d}}x = \left( {\sin x + x\cos x - \sin x} \right)dx = x\cos x\,{\rm{d}}x.$

Suy ra $\int {\dfrac{{x\cos x}}{{x\sin x + \cos x}}{\rm{d}}x}  = \int {\dfrac{{{\rm{d}}t}}{t}}  = \ln \left| t \right| + C = \ln \left| {x\sin x + \cos x} \right| + C.$

Do đó

$\begin{array}{l}F\left( x \right) = \int {f\left( x \right){\rm{d}}x}  = \dfrac{{{x^2}}}{2} + \ln \left| {x\sin x + \cos x} \right| + C.\\ \Rightarrow F\left( 0 \right) = C = 1 \Rightarrow F\left( x \right) = \dfrac{{{x^2}}}{2} + \ln \left| {x\sin x + \cos x} \right| + 1.\\ \Rightarrow F\left( {\dfrac{\pi }{2}} \right) = \dfrac{{{\pi ^2}}}{8} + \ln \dfrac{\pi }{2} + 1.\end{array}$

Ta có: \(F\left( x \right) = \int {f\left( x \right)dx}  = \int {x\sqrt {{x^2} - m} dx} \).

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

\( \Rightarrow F\left( x \right) = \int {t.tdt}  = \int {{t^2}dt}  = \dfrac{{{t^3}}}{3} + C = \dfrac{{{{\left( {\sqrt {{x^2} - m} } \right)}^3}}}{3} + C\).

Theo bài ra ta có: \(\left\{ \begin{array}{l}F\left( {\sqrt 2 } \right) = \dfrac{7}{3}\\F\left( {\sqrt 5 } \right) = \dfrac{{14}}{3}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {2 - m} } \right)}^3}}}{3} + C = \dfrac{7}{3}\\\dfrac{{{{\left( {\sqrt {5 - m} } \right)}^3}}}{3} + C = \dfrac{{14}}{3}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {2 - m} } \right)}^3}}}{3} + C = \dfrac{7}{3}\\\dfrac{{{{\left( {\sqrt {5 - m} } \right)}^3}}}{3} - \dfrac{{{{\left( {\sqrt {2 - m} } \right)}^3}}}{3} = \dfrac{7}{3}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {2 - m} } \right)}^3}}}{3} + C = \dfrac{7}{3}\\{\left( {\sqrt {5 - m} } \right)^3} - {\left( {\sqrt {2 - m} } \right)^3} = 7\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {2 - m} } \right)}^3}}}{3} + C = \dfrac{7}{3}\\{\left( {\sqrt {5 - m} } \right)^3} - {\left( {\sqrt {2 - m} } \right)^3} - 7 = 0\,\,\,\left( * \right)\end{array} \right.\)

Xét hàm số \(f\left( m \right) = {\left( {\sqrt {5 - m} } \right)^3} - {\left( {\sqrt {2 - m} } \right)^3} - 7\) với \(m \le 2\).

Ta có \(f'\left( m \right) =  - \dfrac{3}{2}\sqrt {5 - m}  + \dfrac{3}{2}\sqrt {2 - m}  = \dfrac{3}{2}\left( {\sqrt {2 - m}  - \sqrt {5 - m} } \right)\).

Vì \(2 - m < 5 - m\,\,\forall m \le 2\) \( \Rightarrow \sqrt {2 - m}  < \sqrt {5 - m} \,\,\forall m \le 2\), do đó \(f'\left( m \right) < 0\,\,\forall m \le 2\).

Suy ra hàm số \(f\left( m \right)\) nghịch biến trên \(\left( { - \infty ;2} \right]\).

Khi đó phương trình (*) có nhiều nhất 1 nghiệm, mà \(f\left( 1 \right) = 0\) nên \(m = 1\) là nghiệm duy nhất của phương trình (*).

Vậy có 1 giá trị của \(m\) thỏa mãn yêu cầu bài toán.

\(\int {f\left( {\dfrac{1}{2}x} \right)dx}  = {x^2} + 4x + C\)

\(\begin{array}{l}t = \dfrac{1}{2}x \Rightarrow x = 2t \Rightarrow dx = 2dt\\\int {f\left( t \right)2dt}  = {\left( {2t} \right)^2} + 4.2t + C\\\int {f\left( t \right)dt}  = 2{t^2} + 4t + \dfrac{C}{2}\\ \Rightarrow f\left( t \right) = 4t + 4\\ \Rightarrow f\left( {x - 2} \right) = 4x - 4\\ \Rightarrow \int {f\left( {x - 2} \right)dx}  = \int {\left( {4x - 4} \right)dx} \\ = 2{x^2} - 4x + C\\ \Rightarrow 2a + b = 2.2 + \left( { - 4} \right) = 0\end{array}\)

Ta có: \(f\left( x \right) = \dfrac{x}{{\sqrt {{x^2} + 4} }}\) \( \Rightarrow f'\left( x \right) = \left( {\dfrac{x}{{\sqrt {{x^2} + 4} }}} \right)'\)

\( \Rightarrow f'\left( x \right) = \dfrac{{\sqrt {{x^2} + 4}  - \dfrac{{{x^2}}}{{\sqrt {{x^2} + 4} }}}}{{{x^2} + 4}}\) \( = \dfrac{{{x^2} + 4 - {x^2}}}{{\sqrt {{{\left( {{x^2} + 4} \right)}^3}} }} = \dfrac{4}{{\sqrt {{{\left( {{x^2} + 4} \right)}^3}} }}.\)

\( \Rightarrow g\left( x \right) = \left( {x + 1} \right)f'\left( x \right)\) \( = xf'\left( x \right) + f'\left( x \right).\)

\(\begin{array}{l} \Rightarrow \int {g\left( x \right)dx}  = \int {xf'\left( x \right)dx}  + \int {f'\left( x \right)dx} \\ = \int {\dfrac{{4x}}{{\sqrt {{{\left( {{x^2} + 4} \right)}^3}} }}dx}  + f\left( x \right)dx\\ = \int {\dfrac{{4x}}{{\sqrt {{{\left( {{x^2} + 4} \right)}^3}} }}dx}  + \dfrac{x}{{\sqrt {{x^2} + 4} }}.\end{array}\)

Đặt \(I = \int {\dfrac{{4x}}{{\sqrt {{{\left( {{x^2} + 4} \right)}^3}} }}dx} \)

Đặt \(\sqrt {{x^2} + 4}  = t\)\( \Rightarrow {t^2} = {x^2} + 4\) \( \Rightarrow xdx = tdt\)

\(\begin{array}{l} \Rightarrow I = \int {\dfrac{{4tdt}}{{{t^3}}} = \int {\dfrac{4}{{{t^2}}}dt}  =  - \dfrac{4}{t} + C} \\ \Rightarrow I =  - \dfrac{4}{{\sqrt {{x^2} + 4} }} + C\\ \Rightarrow \int {g\left( x \right)dx}  =  - \dfrac{4}{{\sqrt {{x^2} + 4} }} + \dfrac{x}{{\sqrt {{x^2} + 4} }} + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{x - 4}}{{\sqrt {{x^2} + 4} }} + C.\end{array}\)