Retomo o tópico da substituição de Euler exemplificado nesta entrada.
Na questão How to find out what changes applied to integral?, no MSE, Dracontis pretende saber como se pode calcular o seguinte integral
que, em Maple, se transforma neste:
Adapto da minha resposta o seguinte cálculo.
1. Podemos usar a substituição de Euler 
+%7D%5Cmathrm%7Bd%7Dt.++%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned}I=\displaystyle\int\dfrac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\dfrac{1}{2}\displaystyle\int\dfrac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t. \end{aligned}")
2. Dado que a função integranda é uma fracção racional, podemos decompô-la em fracções parciais e de seguida integrar cada uma dessas fracções
+%7D=1-%5Cdfrac%7B1%7D%7Bt%5E%7B2%7D%7D+++%5Cdfrac%7B4%7D%7Bt%7D+%5Cdfrac%7B20%7D%7Bt%5E%7B2%7D-4t-1%7D.&bg=ffffff&fg=000000&s=0 "\dfrac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }=1-\dfrac{1}{t^{2}}+ \dfrac{4}{t}+\dfrac{20}{t^{2}-4t-1}.")
Cálculo pormenorizado. De 
+%5Cmathrm%7Bd%7Dt%5C%5C&=%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cint%5Cdfrac%7B1+2t%5E%7B2%7D+t%5E%7B4%7D%7D%7Bt%5E%7B2%7D%5Cleft(+-1+t%5E%7B2%7D-4t%5Cright)+%7D%5Cmathrm%7Bd%7Dt.%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned}I&=\displaystyle\int\dfrac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\displaystyle\int\dfrac{t+\dfrac{1-t^{2}}{2t}}{\dfrac{1-t^{2}}{2t}+2}\left( -\dfrac{t^{2}+1}{2t^{2}}\right) \mathrm{d}t\\&=\dfrac{1}{2}\displaystyle\int\dfrac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t.\end{aligned}")
Decompondo nas fracções parciais indicadas acima, obtemos
+%5C%5C&-2%5Csqrt%7B5%7D%5Cln%5Cdfrac%7B%5Csqrt%7B5%7D%5Cleft(+%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%5Cright)+-2%5Csqrt%7B5%7D+5%7D%7B5-%5Csqrt%7B5%7D%5Cleft(+%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%5Cright)++2%5Csqrt%7B5%7D%7D+C.++%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned}2I&=\displaystyle\int 1-\dfrac{1}{t^{2}}+\dfrac{4}{t}+\dfrac{20}{t^{2}-4t-1}\mathrm{d}t\\&=\displaystyle\int 1\mathrm{d}t-\displaystyle\int\dfrac{1}{t^{2}}\mathrm{d}t+4\displaystyle\int\frac{1}{t}\mathrm{d}t+20\displaystyle\int\dfrac{1}{t^{2}-4t-1}\mathrm{d}t\\&=t+\dfrac{1}{t}+4\ln\left\vert t\right\vert -2\sqrt{5}\ln \dfrac{\sqrt{5}t-2\sqrt{5}+5}{5-\sqrt{5}t+2\sqrt{5}}+C\\&=\sqrt{x^{2}+1}-x+\dfrac{1}{\sqrt{x^{2}+1}-x}+4\ln\left( \sqrt{x^{2}+1}-x\right) \\&-2\sqrt{5}\ln\dfrac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{aligned}")
Logo o integral dado é igual a
++%5Cdfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%7D+2%5Cln%5Cleft(+%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%5Cright)%5C%5C&-%5Csqrt%7B5%7D%5Cln%5Cdfrac%7B%5Csqrt%7B5%7D%5Cleft(+%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%5Cright)+-2%5Csqrt%7B5%7D+5%7D%7B5-%5Csqrt%7B5%7D%5Cleft(+%5Csqrt%7Bx%5E%7B2%7D+1%7D-x%5Cright)++2%5Csqrt%7B5%7D%7D+C.++%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned}I&=\dfrac{1}{2}\left( \sqrt{x^{2}+1}-x\right) +\dfrac{1}{2}\frac{1}{\sqrt{x^{2}+1}-x}+2\ln\left( \sqrt{x^{2}+1}-x\right)\\&-\sqrt{5}\ln\dfrac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{aligned}")
Filed under: Cálculo, Integrais, Matemática, Mathematics Stack Exchange Tagged: Matemática, MSE 
Continue a ler
…
…z%5E%7B2%7D+8b%5E%7B2%7Dz-7+b%5E%7B2%7D=0&bg=ffffff&fg=000000&s=0 "z^{3}-(b+6)z^{2}+8b^{2}z-7+b^{2}=0")
, sabendo que 
é uma das raízes.=z%5E%7B3%7D+Az%5E%7B2%7D+Bz+C%5Cqquad+(1),&bg=ffffff&fg=000000&s=0 "P(z)=z^{3}+Az^{2}+Bz+C\qquad (1),")
e 
são números reais. Se designarmos as suas raízes por 
e 
, então o polinómio decompõe-se nos factores&=%5Cleft(+z-z_%7B1%7D%5Cright)+%5Cleft(+z-z_%7B2%7D%5Cright)+(z-z_%7B3%7D)%5C%5C++&=z%5E%7B3%7D-%5Cleft(+z_%7B1%7D+z_%7B3%7D+z_%7B2%7D%5Cright)+z%5E%7B2%7D+%5Cleft(z_%7B1%7Dz_%7B2%7D+z_%7B2%7Dz_%7B3%7D+z_%7B1%7Dz_%7B3%7D%5Cright)+z-z_%7B1%7Dz_%7B2%7Dz_%7B3%7D.%5Cend%7Baligned%7D%5Cqquad+(2)&bg=ffffff&fg=000000&s=0 "\begin{aligned}P(z)&=\left( z-z_{1}\right) \left( z-z_{2}\right) (z-z_{3})\\ &=z^{3}-\left( z_{1}+z_{3}+z_{2}\right) z^{2}+\left(z_{1}z_{2}+z_{2}z_{3}+z_{1}z_{3}\right) z-z_{1}z_{2}z_{3}.\end{aligned}\qquad (2)")
=C=-z_%7B1%7Dz_%7B2%7Dz_%7B3%7D.&bg=ffffff&fg=000000&s=0 "P(0)=C=-z_{1}z_{2}z_{3}.")
&bg=ffffff&fg=000000&s=0 "A=-(b+6)")
, 
e 
. Dado que 
é outra, como concluíu. Por isso tem-se +%5Cleft(+1-i%5Cright)=2&bg=ffffff&fg=000000&s=0 "z_{1}z_{2}=\left( 1+i\right) \left( 1-i\right)=2")
e&bg=ffffff&fg=000000&s=0 "\begin{aligned}-7+b^{2}=-2z_{3}\end{aligned},\qquad (3)")
&bg=ffffff&fg=000000&s=0 "z_{3}=\dfrac{7-b^{2}}{2}.\qquad (4)")
=P(z_2)=0&bg=ffffff&fg=000000&s=0 "P(z_1)=P(z_2)=0")
, tem-se+%5E%7B3%7D-(b+6)%5Cleft(+1+i%5Cright)+%5E%7B2%7D+8b%5E%7B2%7D%5Cleft(+1+i%5Cright)+-7+b%5E%7B2%7D%5C%5C&=-9+9b%5E%7B2%7D+i%5Cleft(+-10-2b+8b%5E%7B2%7D%5Cright)=0,%5Cend%7Baligned%7D%5Cqquad+(5)&bg=ffffff&fg=000000&s=0 "\begin{aligned}&\left( 1+i\right) ^{3}-(b+6)\left( 1+i\right) ^{2}+8b^{2}\left( 1+i\right) -7+b^{2}\\&=-9+9b^{2}+i\left( -10-2b+8b^{2}\right)=0,\end{aligned}\qquad (5)")
+%5E%7B3%7D-(b+6)%5Cleft(+1-i%5Cright)+%5E%7B2%7D+8b%5E%7B2%7D%5Cleft(+1-i%5Cright)++-7+b%5E%7B2%7D%5C%5C&=-9+9b%5E%7B2%7D+i%5Cleft(+10+2b-8b%5E%7B2%7D%5Cright)=0,%5Cend%7Baligned%7D%5Cqquad+(6)&bg=ffffff&fg=000000&s=0 "\begin{aligned}&\left( 1-i\right) ^{3}-(b+6)\left( 1-i\right) ^{2}+8b^{2}\left( 1-i\right) -7+b^{2}\\&=-9+9b^{2}+i\left( 10+2b-8b^{2}\right)=0,\end{aligned}\qquad (6)")
verifica o sistema&bg=ffffff&fg=000000&s=0 "\begin{aligned}\left\{\begin{array}{c}-9+9b^{2}=0\\10+2b-8b^{2}=0.\end{array}\right.\end{aligned}\qquad (7)")
&bg=ffffff&fg=000000&s=0 "(7)")
é 
. Usando &bg=ffffff&fg=000000&s=0 "(4)")
determina-se&bg=ffffff&fg=000000&s=0 "z_{3}=3.\qquad (8)")
…
e 
são os semi-eixos. Assim, tem-se
+=ab%5Cpi+.&bg=ffffff&fg=000000&s=0 "4\displaystyle\int_{0}^{a}b\sqrt{1-\dfrac{x^{2}}{a^{2}}}\; \mathrm{d}x=4\left( \dfrac{1}{2}ba\arcsin 1\right) =ab\pi .")
…
…
é a região de 
definida por%5Cin%5Cmathbb%7BR%7D%5E%7B2%7D:(x-2)%5E%7B2%7D+y%5E%7B2%7D%5Cleq+4%5Cwedge+y%5Cgeq+0%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "R=\left\{ \left( x,y\right)\in\mathbb{R}^{2}:(x-2)^{2}+y^{2}\leq 4\wedge y\geq 0\right\}")
%5Cin%5Cmathbb%7BR%7D%5E%7B2%7D:0%5Cleq+r%5Cleq+4%5Ccos%5Ctheta%5Cwedge+0%5Cleq+%5Ctheta%5Cleq%5Cdfrac%7B%5Cpi+%7D%7B2%7D%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "R=\left\{ (r,\theta )\in\mathbb{R}^{2}:0\leq r\leq 4\cos\theta\wedge 0\leq \theta\leq\dfrac{\pi }{2}\right\}")
…
onde o integral a seguir indicado é convergente+%7D%5C;%5Cmathrm%7Bd%7Dt&bg=ffffff&fg=000000&s=0 "I=\displaystyle\int_{0}^{\infty }\frac{1}{t^{x}\left( 1+t^{y}\right) }\;\mathrm{d}t")
+%7D%5C;%5Cmathrm%7Bd%7Dt+%5Cdisplaystyle%5Cint_%7B1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bt%5E%7Bx%7D%5Cleft(+1+t%5E%7By%7D%5Cright)+%7D%5C;%5Cmathrm%7Bd%7Dt=I_%7B1%7D+I_%7B2%7D&bg=ffffff&fg=000000&s=0 "I=\displaystyle\int_{0}^{1}\dfrac{1}{t^{x}\left( 1+t^{y}\right) }\;\mathrm{d}t+\displaystyle\int_{1}^{\infty }\dfrac{1}{t^{x}\left( 1+t^{y}\right) }\;\mathrm{d}t=I_{1}+I_{2}")
+%7D%5C;%5Cmathrm%7Bd%7Dt%5C%5CI_%7B2%7D+&=&%5Cdisplaystyle%5Cint_%7B1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bt%5E%7Bx%7D%5Cleft(+1+t%5E%7By%7D%5Cright)+%7D%5C;%5Cmathrm%7Bd%7Dt%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned}I_{1}&=&\displaystyle\int_{0}^{1}\dfrac{1}{t^{x}\left( 1+t^{y}\right) }\;\mathrm{d}t\\I_{2} &=&\displaystyle\int_{1}^{\infty }\dfrac{1}{t^{x}\left( 1+t^{y}\right) }\;\mathrm{d}t\end{aligned}")
=1/(t%5E%7Bx%7D%5Cleft(+1+t%5E%7By%7D%5Cright)+)&bg=ffffff&fg=000000&s=0 "f(t)=1/(t^{x}\left( 1+t^{y}\right) )")
e escolha-se a função =1/t%5E%7Bx%7D&bg=ffffff&fg=000000&s=0 "g_{1}(t)=1/t^{x}")
. Aplique-se o critério de convergência do limite. Como%7D%7Bg_%7B1%7D(t)%7D=%5Clim_%7Bt%5Crightarrow+0%5E%7B+%7D%7D%5Cdfrac%7B1%7D%7B1+t%5E%7By%7D%7D=1%5Cqquad+(y%3C0)&bg=ffffff&fg=000000&s=0 "\lim_{t\rightarrow 0^{+}}\dfrac{f(t)}{g_{1}(t)}=\lim_{t\rightarrow 0^{+}}\dfrac{1}{1+t^{y}}=1\qquad (y<0)")
é convergente, se 
, donde o integral 
converge na região +:x%3C1%5Cwedge+y%3E0%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "R_{1}=\left\{ \left( x,y\right) :x<1\wedge y>0\right\}")
. Seja agora =1/t%5E%7Bx+y%7D&bg=ffffff&fg=000000&s=0 "g_{2}(t)=1/t^{x+y}")
. Então, aplicando novamente o critério do limite%7D%7Bg_%7B2%7D(t)%7D=%5Clim_%7Bt%5Crightarrow+%5Cinfty+%7D%5Cdfrac%7Bt%5E%7Bx+y%7D%7D%7Bt%5E%7Bx%7D%5Cleft(+1+t%5E%7By%7D%5Cright)+%7D=%5Clim_%7Bt%5Crightarrow+%5Cinfty+%7D%5Cdfrac%7Bt%5E%7By%7D%7D%7B1+t%5E%7By%7D%7D=1%5Cqquad+(y%3E0)&bg=ffffff&fg=000000&s=0 "\lim_{t\rightarrow \infty }\dfrac{f(t)}{g_{2}(t)}=\lim_{t\rightarrow \infty }\dfrac{t^{x+y}}{t^{x}\left( 1+t^{y}\right) }=\lim_{t\rightarrow \infty }\dfrac{t^{y}}{1+t^{y}}=1\qquad (y>0)")
o integral +%7D%5C;%5Cmathrm%7Bd%7Dt&bg=ffffff&fg=000000&s=0 "\displaystyle\int_{1}^{\infty }t^{-\left( x+y\right) }\;\mathrm{d}t")
é convergente, conclui-se que 
converge na região +:x+y%3E1%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "R_{2}=\left\{ \left( x,y\right) :x+y>1\right\}")
. Resulta por fim que 
converge em+:x%3C1%5Cwedge+y%3E1-x%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "R=R_{1}\cap R_{2}=\left\{ \left( x,y\right) :x<1\wedge y>1-x\right\}")
e acima da recta 
.
…