Question

Calcule a integral indefinida

o)

Answer 1

Oi Lucas

∫ (x^4 + 3x^(-1/2) + 4)/√x dx = 

∫ (x^(7/2) + 3/x + 4x^(-1/2) dx = 2x^(9/2)/9 + 3ln(x) - 8√x + C 

Answer 2

$I=\displaystyle\int\!\dfrac{x^4+3x^{-1/2}+4}{\sqrt{x}}\,dx\\\\\\ =\int\!\dfrac{x^4+3x^{-1/2}+4}{x^{1/2}}\,dx\\\\\\ =\int\!\left(\dfrac{x^4}{x^{1/2}}+\dfrac{3x^{-1/2}}{x^{1/2}}+\dfrac{4}{x^{1/2}} \right )\,dx\\\\\\ =\int\!\left(x^{4-(1/2)}+3x^{-1/2-(1/2)}+4x^{-1/2} \right )\,dx\\\\\\ =\int\!\left(x^{7/2}+3x^{-1}+4x^{-1/2} \right )\,dx~~~~~~\mathbf{(i)}$

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Regra para primitiva de potência:

$\displaystyle\int\!x^n\,dx=\left\{\!\begin{array}{ll} \mathrm{\ell n}\!\left|x\right|\,,&\text{se }n=-1\\\\ \dfrac{x^{n+1}}{n+1}\,,&\text{se }n\ne -1 \end{array}\right.$

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Voltando a $\mathbf{(i)},$ temos

$I=\dfrac{x^{(7/2)+1}}{\frac{7}{2}+1}+3\,\mathrm{\ell n}\!\left|x\right|+4\cdot \dfrac{x^{(-1/2)+1}}{-\frac{1}{2}+1}+C\\\\\\ =\dfrac{x^{9/2}}{\frac{9}{2}}+3\,\mathrm{\ell n}\!\left|x\right|+4\cdot \dfrac{x^{1/2}}{\frac{1}{2}}+C\\\\\\ =\dfrac{2}{9}\,x^{9/2}+3\,\mathrm{\ell n}\!\left|x\right|+4\cdot 2x^{1/2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\dfrac{x^4+3x^{-1/2}+4}{\sqrt{x}}\,dx=\dfrac{2}{9}\,x^{9/2}+3\,\mathrm{\ell n}\!\left|x\right|+8\sqrt{x}+C \end{array}}$

( no caso desta questão, pode-se dispensar o uso do módulo no logaritmo, visto que o domínio da função do integrando é somente os reais positivos )

Bons estudos! :-)