Question

Bom dia ? Integral ? por substituição?$\int\limits {x} \sqrt{x^{2}+1 } \, dx$

Answer 1

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


Fazendo a seguinte substituição:

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


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

$=\dfrac{1}{2}\displaystyle\int\! \sqrt{u}\,du\\\\\\ =\dfrac{1}{2}\int\! u^{1/2}\,du\\\\\\ =\dfrac{1}{2}\cdot \dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =\dfrac{1}{2}\cdot \dfrac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =\dfrac{1}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{3}\,u^{3/2}+C\\\\\\ =\dfrac{1}{3}\,u^{3/2}+C\\\\\\ =\dfrac{1}{3}\,(x^2+1)^{3/2}+C\\\\\\ =\dfrac{1}{3}\sqrt{(x^2+1)^3}+C$