2^{\circ} ) Determine as integrais indefinidas .
\int x^{\frac{3}{2}}\left(x^{3}+5 x-2\right) d x

Passo 1

O enunciado pede para calcular a integral indefinida
\int x^{\frac{3}{2}}\bigl(x^{3} + 5x - 2\bigr)\,dx.
O conteúdo que vamos usar é a regra de potência para integrais:
\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C,
válida para n\neq -1.

Passo 2

Primeiro, expandimos o produto dentro da integral usando as propriedades de potências:
x^{\frac{3}{2}}\cdot x^{3} = x^{\frac{3}{2} + 3} = x^{\frac{9}{2}},\quad x^{\frac{3}{2}}\cdot 5x = 5\,x^{\frac{3}{2} + 1} = 5\,x^{\frac{5}{2}},\quad x^{\frac{3}{2}}\cdot(-2) = -2\,x^{\frac{3}{2}}.
Logo a integral fica
\int \bigl(x^{\tfrac{9}{2}} + 5\,x^{\tfrac{5}{2}} - 2\,x^{\tfrac{3}{2}}\bigr)\,dx.

Passo 3

Integramos cada termo separadamente pela regra de potência:
  1. Para \displaystyle \int x^{\tfrac{9}{2}}\,dx, temos n=\tfrac{9}{2}, logo
\int x^{\tfrac{9}{2}}\,dx = \frac{x^{\tfrac{9}{2}+1}}{\tfrac{9}{2}+1} = \frac{x^{\tfrac{11}{2}}}{\tfrac{11}{2}} = \frac{2}{11}\,x^{\tfrac{11}{2}}.
  1. Para \displaystyle \int 5\,x^{\tfrac{5}{2}}\,dx, fatoramos o 5 e aplicamos a regra com n=\tfrac{5}{2}:
5\int x^{\tfrac{5}{2}}\,dx = 5\cdot\frac{x^{\tfrac{5}{2}+1}}{\tfrac{5}{2}+1} = 5\cdot\frac{x^{\tfrac{7}{2}}}{\tfrac{7}{2}} = 5\cdot\frac{2}{7}\,x^{\tfrac{7}{2}} = \frac{10}{7}\,x^{\tfrac{7}{2}}.
  1. Para \displaystyle \int -2\,x^{\tfrac{3}{2}}\,dx, fatoramos o -2 e usamos n=\tfrac{3}{2}:
-2\int x^{\tfrac{3}{2}}\,dx = -2\cdot\frac{x^{\tfrac{3}{2}+1}}{\tfrac{3}{2}+1} = -2\cdot\frac{x^{\tfrac{5}{2}}}{\tfrac{5}{2}} = -2\cdot\frac{2}{5}\,x^{\tfrac{5}{2}} = -\frac{4}{5}\,x^{\tfrac{5}{2}}.

Passo 4

Somando os três resultados e acrescentando a constante de integração C, obtemos
\int x^{\frac{3}{2}}\bigl(x^{3} + 5x - 2\bigr)\,dx = \frac{2}{11}\,x^{\tfrac{11}{2}} + \frac{10}{7}\,x^{\tfrac{7}{2}} - \frac{4}{5}\,x^{\tfrac{5}{2}} + C.

Resposta

\boxed{ \int x^{\frac{3}{2}}\bigl(x^{3} + 5x - 2\bigr)\,dx = \frac{2}{11}\,x^{\tfrac{11}{2}} + \frac{10}{7}\,x^{\tfrac{7}{2}} - \frac{4}{5}\,x^{\tfrac{5}{2}} + C. }

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