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#1 Zara Asker

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Napisano 18.02.2017 - 21:06

\int \frac{5x^2-2x+10}{ \sqrt{3x^2-5x+8} } dx


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Afroman

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Napisano 25.09.2011 - 17:55

#2 Jarekzulus

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Napisano 22.02.2017 - 13:30


\int \frac{5x^2-2x+10}{ \sqrt{3x^2-5x+8} } dx

 

5 x^{2} - 2 x + 10=\frac{19 x}{3} + \frac{5}{3} \left(3 x^{2} - 5 x + 8\right) - \frac{10}{3}

 

\int{\left(\frac{\frac{19 x}{3} - \frac{10}{3}}{\sqrt{3 x^{2} - 5 x + 8}}+ \frac{5}{3} \sqrt{3 x^{2} - 5 x + 8}\right)d x = \int{\frac{\frac{19 x}{3} - \frac{10}{3}}{\sqrt{3 x^{2} - 5 x + 8}} d x} + \int{\frac{5}{3} \sqrt{3 x^{2} - 5 x + 8} d x}

 

Najpierw może ta druga - robisz przekształcenie

 

3 x^{2} - 5 x + 8 = \left(\sqrt{3} x - \frac{5 \sqrt{3}}{6}\right)^{2} + \frac{71}{12}

 

\int \sqrt{3\:x^2\:-\:5\:x\:+\:8}=\int \sqrt{\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)^2+\frac{71}{12}}dx

 

i podstawienie u=\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)\quad \:du=\sqrt{3}dx

 

=\int \frac{1}{6}\sqrt{12u^2+71}du=\[=\frac{\sqrt{71}}{2\sqrt{3}}tg \left(v\right)\quad \:du=\frac{\sqrt{71}}{2\sqrt{3}}\sec ^2\left(v\right)dv\]=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\cdot \int \:\sqrt{71tg ^2\left(v\right)+71}\sec ^2\left(v\right)dv

 

=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\cdot \int \:\sqrt{71}\sqrt{tg ^2\left(v\right)+1}\sec ^2\left(v\right)dv=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\cdot \int \:\sec ^2\left(v\right)\sqrt{\sec ^2\left(v\right)}\sqrt{71}dv=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\cdot \int \:\sqrt{71}\sec ^3\left(v\right)dv

 

wykorzystując wzór

 

\re \int \:\sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx

 

mamy

 

=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\sqrt{71}\left(\frac{\sec ^2\left(v\right)\sin \left(v\right)}{2}+\frac{1}{2}\cdot \int \:\sec \left(v\right)dv\right)

 

\re \int \:\sec \left(v\right)dv=\ln \left(tg\left(v\right)+\sec \left(v\right)\right)

 

=\frac{1}{6}\cdot \frac{1}{2}\sqrt{\frac{71}{3}}\sqrt{71}\left(\frac{\sec ^2\left(v\right)\sin \left(v\right)}{2}+\frac{1}{2}\ln \left(tg \left(v\right)+\sec \left(v\right)\right)\right)

 

pamiętaj, że                  v=arctg \left(\frac{2\sqrt{3}}{\sqrt{71}}u\right)                u=\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)

 

A teraz druga... znaczy pierwsza

 

\int \frac{\frac{19 x}{3} - \frac{10}{3}}{\sqrt{3 x^{2} - 5 x + 8}} d x

 

\int \frac{\frac{19 x}{3} - \frac{10}{3}}{\sqrt{3 x^{2} - 5 x + 8}} d x=\int \frac{\frac{19x}{3}-\frac{10}{3}}{\sqrt{\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)^2+\frac{71}{12}}}dx\[u=\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)\quad \:du=\sqrt{3}dx\]=\int \frac{2\left(19\left(\frac{u}{\sqrt{3}}+\frac{5}{6}\right)-10\right)}{3\sqrt{12u^2+71}}du

 

jak poskracasz =\frac{2}{3}\cdot \int \:\frac{38\sqrt{3}u+35}{6\sqrt{12u^2+71}}du=\frac{2}{3}\cdot \frac{1}{6}\left(\int \frac{38\sqrt{3}u}{\sqrt{12u^2+71}}du+\int \frac{35}{\sqrt{12u^2+71}}du\right)

 

\int \frac{38\sqrt{3}u}{\sqrt{12u^2+71}}du=38\sqrt{3}\cdot \int \:\frac{u}{\sqrt{12u^2+71}}du=\[v=12u^2+71\quad \:dv=24udu\]=38\sqrt{3}\frac{1}{24}\cdot \int \:\frac{1}{\sqrt{v}}dv=\frac{19\sqrt{12u^2+71}}{2\sqrt{3}}+C

 

\int \frac{35}{\sqrt{12u^2+71}}du=35\cdot \int \:\frac{1}{\sqrt{12u^2+71}}du=\[u=\frac{\sqrt{71}}{2\sqrt{3}}tg \left(v\right)\quad \:du=\frac{\sqrt{71}}{2\sqrt{3}}\sec ^2\left(v\right)dv\]=35\cdot \int \:\frac{\sqrt{\frac{71}{3}}\sec ^2\left(v\right)}{2\sqrt{71tg^2\left(v\right)+71}}dv

 

=35\cdot \frac{\sqrt{\frac{71}{3}}}{2}\cdot \int \:\frac{\sec ^2\left(v\right)}{\sqrt{71}\sqrt{tg^2\left(v\right)+1}}dv=35\cdot \frac{\sqrt{\frac{71}{3}}}{2}\cdot \int \:\frac{\sec ^2\left(v\right)}{\sqrt{\sec ^2\left(v\right)}\sqrt{71}}dv=35\cdot \frac{\sqrt{\frac{71}{3}}}{2}\cdot \frac{1}{\sqrt{71}}\cdot \int \:\sec \left(v\right)dv

 

\int \:\sec \left(v\right)dv=\ln \left(tg \left(v\right)+\sec \left(v\right)\right)

 

=35\cdot \frac{\sqrt{\frac{71}{3}}}{2}\cdot \frac{1}{\sqrt{71}}\ln \left(tg \left(v\right)+\sec \left(v\right)\right)+C

 

\[v=arctg \left(\frac{2\sqrt{3}}{\sqrt{71}}u\right)\]=35\cdot \frac{\sqrt{\frac{71}{3}}}{2}\cdot \frac{1}{\sqrt{71}}\ln \left(tg \left(arctg \left(\frac{2\sqrt{3}}{\sqrt{71}}u\right)\right)+\sec \left(arctg \left(\frac{2\sqrt{3}}{\sqrt{71}}u\right)\right)\right)=\frac{35\ln \left(\sqrt{\frac{12u^2}{71}+1}+2\sqrt{\frac{3}{71}}u\right)}{2\sqrt{3}}

 

czyli

 

\int \frac{\frac{19x}{3}-\frac{10}{3}}{\sqrt{3x^2-5x+8}}dx=\frac{2}{3}\cdot \frac{1}{6}\left(\int \frac{38\sqrt{3}u}{\sqrt{12u^2+71}}du+\int \frac{35}{\sqrt{12u^2+71}}du\right)=\frac{2}{3}\cdot \frac{1}{6}\left(\frac{19\sqrt{12u^2+71}}{2\sqrt{3}}+\frac{35\ln \left(\sqrt{\frac{12u^2}{71}+1}+2\sqrt{\frac{3}{71}}u\right)}{2\sqrt{3}}\right)

 

u=\left(\sqrt{3}x-\frac{5}{2\sqrt{3}}\right)

 

Można chyba dużo szybciej ;)


Użytkownik Jarekzulus edytował ten post 22.02.2017 - 13:34

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:wave: :wave: :wave: Jeśli rzuciłem choć promyczek światła na problem który postawiłeś - podziękuj. pre_1433974176__syg.jpgNad kreską


#3 Mariusz M

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Napisano 04.04.2017 - 20:03

Dosyć dobry efekt daje pierwsze podstawienie Eulera ale jak to modnie jest unikać tych podstawień 

 

\int{\frac{5x^2-2x+10}{\sqrt{3x^2-5x+8}}\mbox{d}x}\\</p>\\<p>\sqrt{3}\int{\frac{5x^2-2x+10}{\sqrt{9x^2-15x+24}}\mbox{d}x}\\</p>\\<p>\sqrt{9x^2-15x+24}=t-3x\\</p>\\<p>9x^2-15x+24=t^2-6tx+9x^2</p>\\<p>-15x+24=t^2-6tx\\</p>\\<p>6tx-15x=t^2-24\\</p>\\<p>x\left(6t-15\right)=t^2-24\\</p>\\<p>x=\frac{t^2-24}{6t-15}\\</p>\\<p>t-3x=\frac{6t^2-15t-3t^2+72}{6t-15}=\frac{3t^2-15t+72}{6t-15}\\</p>\\<p>\mbox{d}x=\frac{2t\left(6t-15\right)-6\left(t^2-24\right)}{\left(6t-15\right)^2}\mbox{d}t\\</p>\\<p>\mbox{d}x=\frac{6t^2-30t+144}{\left(6t-15\right)^2}\mbox{d}t\\</p>\\<p>\sqrt{3}\int{\left(5\frac{\left(t^2-24\right)^2}{\left(6t-15\right)^2}-2\frac{t^2-24}{6t-15}+10\right)\frac{6t-15}{3t^2-15t+72}\cdot\frac{2\left(3t^2-15t+72\right)}{\left(6t-15\right)^2}\mbox{d}t}\\</p>\\<p>\sqrt{3}\int{\left(\frac{5\left(t^4-48t^2+576\right)-2\left(t^2-24\right)\left(6t-15\right)+10\left(36t^2-180t+225\right)}{\left(6t-15\right)^2}\right)\cdot\frac{2}{6t-15}\mbox{d}t}\\</p>\\<p>2\sqrt{3}\int{\frac{5t^4-240t^2+2880-2\left(6t^3-15t^2-144t+360\right)+360t^2-1800t+2250}{\left(6t-15\right)^3}\mbox{d}t}\\</p>\\<p>2\sqrt{3}\int{\frac{5t^4-240t^2+2880-12t^3+30t^2+288t-720+360t^2-1800t+2250}{\left(6t-15\right)^3}\mbox{d}t}\\</p>\\<p>2\sqrt{3}\int{\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(6t-15\right)^3}\mbox{d}t}\\</p>\\<p>\frac{2\sqrt{3}}{27}\int{\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^3}\mbox{d}t}\\</p>\\<p>-\frac{\sqrt{3}}{54}\int{\left(5t^4-12t^3+150t^2-1512t+4410\right)\frac{\left(-4\right)}{\left(2t-5\right)^3}\mbox{d}t}\\</p>\\<p>-\frac{\sqrt{3}}{54}\left(\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^2}-\int{\frac{20t^3-36t^2+300t-1512}{\left(2t-5\right)^2}\mbox{d}t}\right)\\</p>\\<p>-\frac{\sqrt{3}}{54}\left(\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^2}+\int{\left(10t^3-18t^2+150t-756\right)\frac{\left(-2\right)}{\left(2t-5\right)^2}\mbox{d}t}\right)\\</p>\\<p>-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-\int{\frac{60t^2-72t+300}{2t-5}\mbox{d}t}\right)\\</p>\\<p>-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-\int{\left(30t+39\right)\mbox{d}t}-\int{\frac{495}{2t-5}\mbox{d}t}\right)\\</p>\\<p>-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-15t^2-39t-\frac{495}{2}\ln{\left|2t-5\right|}\right)+C\\</p>\\<p>


Użytkownik Mariusz M edytował ten post 04.04.2017 - 21:15

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