Me ajuda a calcular a integral PORFAVOOR ♥
Anexos:
Soluções para a tarefa
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Boa noite Marly!
Solução!
Primeiro vamos reescrever a integral!
![\displaystyle \int \frac{2 \sqrt{x} }{4x}+ \frac{x}{4x}dx\\\\\\\\\\\
\dfrac{1}{4} \displaystyle \int \frac{2 \sqrt{x} }{x}+ \dfrac{x}{x}dx\\\\\\\\\
\dfrac{1}{4} \displaystyle \int \frac{2 \sqrt{x} }{x}+ 1dx\\\\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint+%5Cfrac%7B2+%5Csqrt%7Bx%7D+%7D%7B4x%7D%2B+%5Cfrac%7Bx%7D%7B4x%7Ddx%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%0A+%5Cdfrac%7B1%7D%7B4%7D+%5Cdisplaystyle+%5Cint+%5Cfrac%7B2+%5Csqrt%7Bx%7D+%7D%7Bx%7D%2B+%5Cdfrac%7Bx%7D%7Bx%7Ddx%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B1%7D%7B4%7D+%5Cdisplaystyle+%5Cint+%5Cfrac%7B2+%5Csqrt%7Bx%7D+%7D%7Bx%7D%2B+1dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+%0A+ "\displaystyle \int \frac{2 \sqrt{x} }{4x}+ \frac{x}{4x}dx\\\\\\\\\\\
\dfrac{1}{4} \displaystyle \int \frac{2 \sqrt{x} }{x}+ \dfrac{x}{x}dx\\\\\\\\\
\dfrac{1}{4} \displaystyle \int \frac{2 \sqrt{x} }{x}+ 1dx\\\\\\\\\")
![\dfrac{1}{4} \displaystyle \int 2x^{ \frac{1}{2} }.x^{-1} + 1dx\\\\\\\\\
\dfrac{1}{4} \displaystyle \int2 x^{ -\frac{1}{2} }) + 1dx\\\\\\\\\
I= \dfrac{1}{4}( \frac{ 2x^{ -\frac{1}{2} +1}}{ -\frac{1}{2} +1} )+1\\\\\\\\
I= \dfrac{1}{4}( \frac{ 2x^{ \frac{1}{2}}}{ \frac{1}{2} } )+x\\\\\\\\
I= \dfrac{1}{4}( 4 \sqrt{x} )+x\\\\\\\\
I= \frac{2 \sqrt{x} +x}{4}+c](https://tex.z-dn.net/?f=+%5Cdfrac%7B1%7D%7B4%7D+%5Cdisplaystyle+%5Cint+2x%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D.x%5E%7B-1%7D+++%2B+1dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B1%7D%7B4%7D+%5Cdisplaystyle+%5Cint2+x%5E%7B+-%5Cfrac%7B1%7D%7B2%7D+%7D%29+%2B+1dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cdfrac%7B1%7D%7B4%7D%28+%5Cfrac%7B+2x%5E%7B+-%5Cfrac%7B1%7D%7B2%7D+%2B1%7D%7D%7B+-%5Cfrac%7B1%7D%7B2%7D+%2B1%7D+%29%2B1%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cdfrac%7B1%7D%7B4%7D%28+%5Cfrac%7B+2x%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%29%2Bx%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cdfrac%7B1%7D%7B4%7D%28+4+%5Csqrt%7Bx%7D+%29%2Bx%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cfrac%7B2+%5Csqrt%7Bx%7D+%2Bx%7D%7B4%7D%2Bc+%0A%0A "\dfrac{1}{4} \displaystyle \int 2x^{ \frac{1}{2} }.x^{-1} + 1dx\\\\\\\\\
\dfrac{1}{4} \displaystyle \int2 x^{ -\frac{1}{2} }) + 1dx\\\\\\\\\
I= \dfrac{1}{4}( \frac{ 2x^{ -\frac{1}{2} +1}}{ -\frac{1}{2} +1} )+1\\\\\\\\
I= \dfrac{1}{4}( \frac{ 2x^{ \frac{1}{2}}}{ \frac{1}{2} } )+x\\\\\\\\
I= \dfrac{1}{4}( 4 \sqrt{x} )+x\\\\\\\\
I= \frac{2 \sqrt{x} +x}{4}+c")
Boa noite!
Bons estudos!
Solução!
Primeiro vamos reescrever a integral!
Boa noite!
Bons estudos!
albertrieben:
um pequeno erro. 1/4*2x^(1/2)/(1/2) = 4/4x^(1/2) = √x
solução √x + x/4 + C
Respondido por
1
Oi Marlyn
∫ (2√x + x)/(4x) dx = 1/2 ∫ 1/√x + ∫ 1/4 dx
∫ 1/√x = 2√x
∫ (2√x + x)/(4x) dx = √x + x/4 + C
.
∫ (2√x + x)/(4x) dx = 1/2 ∫ 1/√x + ∫ 1/4 dx
∫ 1/√x = 2√x
∫ (2√x + x)/(4x) dx = √x + x/4 + C
.
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