Question

Calcule a integral por substituição de variável

14) $\int\ {} \frac{x^2}{ \sqrt{x^3+1} } \, dx$

Answer 1

$\displaystyle I=\int\!\frac{x^2}{\sqrt{x^3+1}}\,dx\\\\\\ =\int\!\frac{1}{3}\cdot 3\cdot \frac{x^2}{\sqrt{x^3+1}}\,dx\\\\\\ =\frac{1}{3}\int\!\frac{1}{\sqrt{x^3+1}}\cdot 3x^2\,dx~~~~~~\mathbf{(i)}$


Faça a seguinte substituição:

$x^3+1=u~~\Rightarrow~~3x^2\,dx=du$


Substituindo, a integral $\mathbf{(i)}$ fica

$\displaystyle=\frac{1}{3}\int\!\frac{1}{\sqrt{u}}\,du\\\\\\ =\frac{1}{3}\int\!\frac{1}{u^{1/2}}\,du\\\\\\ =\frac{1}{3}\int\!u^{-1/2}\,du\\\\\\ =\frac{1}{3}\cdot \frac{u^{(-1/2)+1}}{-\frac{1}{2}+1}+C\\\\\\ =\frac{1}{3}\cdot \frac{u^{1/2}}{\frac{1}{2}}+C\\\\\\ =\frac{1}{3}\cdot 2\,u^{1/2}+C\\\\\\ =\frac{2}{3}\sqrt{u}+C\\\\\\ =\frac{2}{3}\sqrt{x^3+1}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{x^2}{\sqrt{x^3+1}}\,dx=\frac{2}{3}\sqrt{x^3+1}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{x^2}{\sqrt{x^3+1}}dx\\\\\\=\int \frac{1}{3\sqrt{u}}du\\\\\\=\frac{1}{3}\cdot \int \frac{1}{\sqrt{u}}du\\\\\\=\frac{1}{3}\cdot \int \frac{1}{u^{\frac{1}{2}}}du\\\\\\=\frac{1}{3}\cdot \int \:u^{-\frac{1}{2}}du\\\\\\=\frac{1}{3}\cdot \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\\\\\\=\frac{1}{3}\cdot \frac{\left(x^3+1\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\\\\\\=\frac{2}{3}\sqrt{x^3+1}$

$\to \boxed{\sf =\frac{2}{3}\sqrt{x^3+1}+C}$