Question

Calcule as integrais indefinidas:

$\int\limits \sqrt{x} (x+\frac{1}{\sqrt{x} } )^{2} dx$

Answer 1

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\sqrt{x}\bigg(x+\dfrac{1}{\sqrt{x}}\bigg)^2\,dx=\int\sqrt{x}\bigg(x^2+\dfrac{2x}{\sqrt{x}}+\dfrac{1}{x}\bigg)\,dx\\\\\displaystyle\rm\int(x^{\frac{5}{2}}+2x+x^{-\frac{1}{2}})\,dx=\dfrac{2}{7}x^{\frac{7}{2}}+x^2+2x^{\frac{1}{2}}+k\end{array}}$

Answer 2

Resposta:

∫ √x *(x+1/√x)²  dx

∫ √x *(x²+ 2x/√x +1/x)  dx

∫ x²√x + 2x +1/√x  dx

∫ √(x⁴*x) + 2x +1/√x  dx

∫ x^5/2 + 2x +x^(-1/2)  dx

x^(5/2+1) / (5/2+1) +2*x²/2  +x^(-1/2+1)/(-1/2+1)  + c

x^(7/2) / (7/2) +x²  +x^(1/2)/(1/2)  + c

(2/7)* x^(7/2) +x² +2*x^(1/2) +c

(2/7)* √x⁷ +x² +2*√x +c