Question

(50 PONTOS) Calcule a integral indefinida:
$\displaystyle\int\!\frac{1}{\sqrt{x}+\,^{3}\!\!\!\sqrt{x}}\,dx$

Answer 1

Integral Racionalizável: 

Resultado é:

∫dx/√x+∛x = 6x^(1/6) - 3∛x + 2√x - 6 ln|1 + x^(1/6)|

Para maiores detalhes verifique o anexo.

Obrigado pela oportunidade.

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05/10/2016
Sepauto 
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Answer 2

$\int \frac{1}{ \sqrt{x} + \sqrt[3]{x} } dx$

$faça \:k=\sqrt[6]{x} \rightarrow\:x={k}^{6} \\ dx=6{k}^{5}dk$

$\int \frac{1}{ \sqrt{x} + \sqrt[3]{x} } dx = \int \frac{6 {k}^{5}}{ {k}^{3} + {k}^{2} }dk \\ \int \frac{6 \cancel{{{k}^{5} }}}{ \cancel{k}^{2}(k+ 1)}dk$

6k³+0k² | k+1

-6k³-6k² 6k²--6k+6

-6k²+0k

+6k²+6k

6k+0

-6k-6

-6

$\int \frac{6 {k}^{3} }{k + 1} \\ =\int (6 {k}^{2} - 6k + 6 - \frac{6}{k + 1})dk \\ 2 {k}^{3} - 3 {k}^{2} + 6k - 6 ln(k + 1) + c$

$\color{blue}{\int \frac{1}{ \sqrt{x} + \sqrt[3]{x} } dx }\\ \color{blue}{2 {( \sqrt[6]{x})}^{3} - 3 {( \sqrt[6]{x} )}^{2} + 6 \sqrt[6]{x}}\\\color{blue}{ - 6 ln( \sqrt[6]{x} + 1 ) + c} \\ \color{blue}{= 2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x}} \\\color{blue}{ - ln( \sqrt[6]{x} + 1 ) + c}$