2^{\circ} ) Determine as integrais indefinidas .
\int\left(x^{4}-x^{\frac{2}{5}}+2 x^{\frac{1}{2}}+3\right) d x

Passo 1

O enunciado pede a integral indefinida \displaystyle \int\bigl(x^{4}-x^{\tfrac{2}{5}}+2x^{\tfrac12}+3\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,\quad n\neq -1,
e a linearidade do operador integral.

Passo 2

Pela linearidade da integral, separamos cada termo:
\int\bigl(x^{4}-x^{\tfrac{2}{5}}+2x^{\tfrac12}+3\bigr)\,dx =\int x^{4}\,dx \;-\;\int x^{\tfrac{2}{5}}\,dx \;+\;2\int x^{\tfrac12}\,dx \;+\;3\int 1\,dx.

Passo 3

Aplicando a regra de potência \int x^{n}\,dx = x^{n+1}/(n+1): \displaystyle\int x^{4}\,dx = \frac{x^{5}}{5}, \displaystyle\int x^{\tfrac{2}{5}}\,dx = \frac{x^{\tfrac{2}{5}+1}}{\tfrac{2}{5}+1} = \frac{x^{\tfrac{7}{5}}}{\tfrac{7}{5}} = \frac{5}{7}x^{\tfrac{7}{5}}, \displaystyle\int x^{\tfrac12}\,dx = \frac{x^{\tfrac32}}{\tfrac32} = \frac{2}{3}x^{\tfrac32}, \displaystyle\int 1\,dx = x.

Passo 4

Substituindo esses resultados na soma e multiplicando constantes:
\int\bigl(x^{4}-x^{\tfrac{2}{5}}+2x^{\tfrac12}+3\bigr)\,dx = \frac{x^{5}}{5} - \frac{5}{7}x^{\tfrac{7}{5}} + 2\!\cdot\!\frac{2}{3}x^{\tfrac32} + 3x + C = \frac{x^{5}}{5} - \frac{5}{7}x^{\tfrac{7}{5}} + \frac{4}{3}x^{\tfrac32} + 3x + C.

Resposta

\boxed{\int\bigl(x^{4}-x^{\tfrac{2}{5}}+2x^{\tfrac12}+3\bigr)\,dx = \frac{x^{5}}{5} - \frac{5}{7}x^{\tfrac{7}{5}} + \frac{4}{3}x^{\tfrac32} + 3x + C.}

Exercícios relacionados