Como calcular a integral abaixo?

 ⌡xˆ2 ⎷x+1 dx

xˆ2⎷x+1, substituir por u=⎷x+1 e du=1/(2⎷x+1) dx

=2⎰uˆ2(uˆ2-1)ˆ2 du

Expandindo: 2⎰(uˆ6-2uˆ4+uˆ2)du

Integração por termos:

=2⎰uˆ6du - 4⎰uˆ4du +2⎰uˆ2du

Integral de uˆ6= (uˆ7)/7

Integral de uˆ4= (uˆ5)/5

Integral de uˆ2= (uˆ3)/3

⎰2uˆ7/7 - 4uˆ5/5+2uˆ3/3+constante

2/7(x+1)ˆ7/2 - 4/5(x+1)ˆ5/2 + 2/3(x+1)ˆ3/2+constante

2/105 (x+1)ˆ3/2 (15xˆ2 - 12x + 8) + constante. 

Obrigadaaa!!!!

Para encontrar a integral dada, realizaremos os cálculos abaixo:

\(\begin{align} & f={{x}^{2}}\sqrt{x+1} \\ & \int_{{}}^{{}}{f(x)=}\int_{{}}^{{}}{{{x}^{2}}\sqrt{x+1}} \\ & u=\sqrt{x+1} \\ & x-{{u}^{2}}-1 \\ & dx=2udu \\ & \int_{{}}^{{}}{{{x}^{2}}\sqrt{x+1}}=\int_{{}}^{{}}{({{u}^{2}}-1)u2udu} \\ & \int_{{}}^{{}}{{{x}^{2}}\sqrt{x+1}}=2\int_{{}}^{{}}{{{u}^{2}}{{({{u}^{2}}-1)}^{2}}du} \\ & \int_{{}}^{{}}{{{x}^{2}}\sqrt{x+1}}=\frac{2{{u}^{7}}}{7}-\frac{4{{u}^{5}}}{5}+\frac{2{{u}^{3}}}{3} \\ & \int_{{}}^{{}}{{{x}^{2}}\sqrt{x+1}}=\frac{2{{(x+1)}^{7/2}}}{7}-\frac{4{{(x+1)}^{5/2}}}{5}+\frac{2{{(x+1)}^{3/2}}}{3}+C \\ \end{align} \)

Portanto,o valor da integral dada será: \(\boxed{\int_{}^{} {{x^2}\sqrt {x + 1} } = \frac{{2{{(x + 1)}^{7/2}}}}{7} - \frac{{4{{(x + 1)}^{5/2}}}}{5} + \frac{{2{{(x + 1)}^{3/2}}}}{3} + C}\).