Question

Calcular a integral de x.sqrt(1+4x²).

Answer 1

$\displaystyle\int x\sqrt{1+4x^2}\,dx\\\\\\ =\int \dfrac{1}{8}\cdot 8x\sqrt{1+4x^2}\,dx\\\\\\ =\dfrac{1}{8}\int \sqrt{1+4x^2}\cdot 8x\,dx~~~~~~\mathbf{(i)}$


Substituição:

$1+4x^2=u~~\Rightarrow~~8x\,dx=du$


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

$=\displaystyle\frac{1}{8}\int \sqrt{u}\,du\\\\\\ =\frac{1}{8}\int u^{1/2}\,du\\\\\\ =\frac{1}{8}\cdot \dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =\frac{1}{8}\cdot \dfrac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =\frac{1}{8}\cdot \dfrac{2}{3}\,u^{3/2}+C\\\\\\ =\frac{1}{12}\,u^{3/2}+C\\\\\\ =\frac{1}{12}\big(1+4x^2\big)^{3/2}+C$

$\therefore~~\boxed{\begin{array}{c}\displaystyle\int x\sqrt{1+4x^2}\,dx=\frac{1}{12}\big(1+4x^2\big)^{3/2}+C \end{array}}$


Bons estudos! :-)