Question

Calcule, pelo método de substituição, a integral indefinida raiz 2 x3 + 4. X2 dx.

Answer 1

$\displaystyle\mathsf{\int \sqrt{2{x}^{3}+4}~{x}^{2}dx=\dfrac{1}{6}\int\sqrt{2{x}^{3}+4}~6{x}^{2}dx}$

faça

$\mathsf{u=2{x}^{3}+4\to~du=6{x}^{2}dx}$

$\displaystyle\mathsf{\dfrac{1}{6}\int\sqrt{{2{x}^{3}+4}}~6{x}^{2}dx=\dfrac{1}{6}\int\sqrt{u}\,du}\\\displaystyle\mathsf{\dfrac{1}{6}.\dfrac{2}{3}\,{u}^{\frac{3}{2}}+k}\\\displaystyle\mathsf{\dfrac{1}{9}\,{u}^{\frac{3}{2}}+k}$

$\boxed{\boxed{\displaystyle\mathsf{\int\,\sqrt{2{x}^{3}+4}~{x}^{2}dx=\dfrac{1}{9}\sqrt{{(2{x}^{3}+4)}^{3}}+k}}}$