Oblicz całkę (x^2=1)/sqrt(3x-1)dx.... Question from @Shameika

December 2018 1 35 Report

Oblicz całkę (x^2=1)/sqrt(3x-1)dx

Paawełek Poprzez podstawianie:

$\displaystyle{} \int \frac{x^2+1}{\sqrt{3x-1}}\mathrm{d}x=\int \frac{x^2}{\sqrt{3x-1}}+\int \frac{\mathrm{d}x}{\sqrt{3x-1}}\\ \\ \hbox{Metoda podstawiania:} \\ \\ t=\sqrt{3x-1} \qquad (...)^2 \\ \\ 3x-1=t^2 \\ 3x=t^2+1 \qquad /:3 \\ \\ x=\frac{t^2+1}{3} \qquad \Rightarrow \qquad x^2= \left(\dfrac{t^2+1}{3}\right)^2 \\ \\ \hbox{Rozniczkujac:} \\ \\ \mathrm{d}x= \frac{2}{3}t\mathrm{d}t$
Jak wszystko podstawisz do całki otrzymasz:

$\displaystyle{} \int \frac{\left(\frac{t^2+1}{3}\right)^2 \cdot \frac{2}{3}t}{t}\mathrm{d}t+ \int \frac{\frac{2}{3}t}{t}= \int \frac{t^4+2t^2+1}{9} \cdot \frac{2}{3}\mathrm{d}t +\int \frac{2}{3}\mathrm{d}t= \\ \\ = \frac{2}{27} \int (t^4+2t^2+1)\mathrm{d}t+\int \frac{2}{3}\mathrm{d}t= \\ \\ = \frac{2}{27}\left(\frac{1}{5}t^5+\frac{2}{3}t^3+t\right)+\frac{2}{3}t+C= \\ \\$

$\displaystyle{} = \frac{2}{135}t^5+\frac{4}{81}t^3+\frac{2}{27}t+\frac{2}{3}t+C=\frac{2}{135}t^5+\frac{4}{81}t^3+\frac{20}{27}t+C= \\ \\ =\frac{2}{405}t (3t^4+10t^2+150)+C \\ \\ \hbox{Powracam do podstawienia:} \\ \\ \frac{2}{405} \sqrt{3x-1}\left(3 \sqrt{3x-1}^4 + 10\sqrt{3x-1}^2+150\right)= \\ \\ \frac{2}{405}\sqrt{3x-1} \left( 3(3x-1)^2 + 10(3x-1) + 150 \right)= \\ \\ = \frac{2}{405}\sqrt{3x-1} (27x^2-18x+3+30x-10+150) = \\ \\ \frac{2}{405}\sqrt{3x-1}(27x^2+12x+143)$

Uhh, wymordowałem tę całkę :) Współczuję Ci i pozdrawiam....

Wynik:

$\displaystyle{} \int \frac{x^2+1}{\sqrt{3x-1}} \mathrm{d}x=\frac{2}{405}\sqrt{3x-1}(27x^2+12x+143)+C$

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$\dfrac{1}{\sqrt{x-5}+\sqrt{x-7}} = \dfrac{\sqrt{x-5}-\sqrt{x-7}}{(\sqrt{x-5}+\sqrt{x-7})(\sqrt{x-5}-\sqrt{x-7})}= \\ \\ = \dfrac{\sqrt{x-5}-\sqrt{x-7}}{x-5-x+7}=\dfrac{\sqrt{x-5}-\sqrt{x-7}}{2}= \\ \\ = \dfrac{1}{2}(x-5)^{\frac{1}{2}} - \dfrac{1}{2}(x-7)^{\frac{1}{2}} \\ \\ \hbox{Pochodne z x-5 i x-7 wynosza 1, wiec wykorzystam wzor:} \\ \\ \displaystyle{} \int x^a \mathrm{d}x=\frac{1}{a+1}x^{a+1}+C$

Stąd Twoja całka wyniesie:

$\displaystyle{} \int \frac{\mathrm{d}x}{\sqrt{x-5}+\sqrt{x-7}}=\frac{1}{2} \int (x-5)^{\frac{1}{2}}\mathrm{d}x -\frac{1}{2} \int (x-7)^{\frac{1}{2}}\mathrm{d}x= \\ \\ = \dfrac{1}{2} \cdot \frac{1}{1+\frac{1}{2}} (x-5)^{1+\frac{1}{2}}-\frac{1}{2} \cdot \frac{1}{1+\frac{1}{2}}(x-7)^{1+\frac{1}{2}}+C= \\ \\ =\frac{1}{3} \left( (x-5)^{\frac{3}{2}} - (x-7)^{\frac{3}{2}}\right)+C$

2 votes Thanks 1