\int_{1}^{2}\left(\sqrt{x}+\frac{1}{\sqrt[3]{x}}+\sqrt[4]{x}\right) d x

Passo 1

O enunciado pede para calcular a integral definida
\int_{1}^{2}\Bigl(\sqrt{x}+\frac{1}{\sqrt[3]{x}}+\sqrt[4]{x}\Bigr)\,dx
Aqui o conteúdo é integração de funções potenciais de x. Vamos usar a regra
\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C,
válida para n\neq -1.

Passo 2

Convertemos cada termo para potência de x:
\sqrt{x}=x^{\tfrac12},\quad \frac{1}{\sqrt[3]{x}}=x^{-\tfrac13},\quad \sqrt[4]{x}=x^{\tfrac14}.
A integral se separa em três pedaços:
\int_{1}^{2}x^{\tfrac12}\,dx +\int_{1}^{2}x^{-\tfrac13}\,dx +\int_{1}^{2}x^{\tfrac14}\,dx.

Passo 3

Aplicando a regra de potências a cada parte:
\int x^{\tfrac12}\,dx =\frac{x^{\tfrac32}}{\tfrac32} =\frac{2}{3}\,x^{\tfrac32},
\int x^{-\tfrac13}\,dx =\frac{x^{\tfrac23}}{\tfrac23} =\frac{3}{2}\,x^{\tfrac23},
\int x^{\tfrac14}\,dx =\frac{x^{\tfrac54}}{\tfrac54} =\frac{4}{5}\,x^{\tfrac54}.
Logo uma antiderivada é
F(x)=\frac{2}{3}x^{\tfrac32}+\frac{3}{2}x^{\tfrac23}+\frac{4}{5}x^{\tfrac54}.

Passo 4

Avaliamo-la de 1 a 2:
F(2)=\frac{2}{3}\,2^{\tfrac32}+\frac{3}{2}\,2^{\tfrac23}+\frac{4}{5}\,2^{\tfrac54} =\frac{2}{3}\cdot2\sqrt{2}+\frac{3}{2}2^{\tfrac23}+\frac{4}{5}\cdot2^{\tfrac54},
F(1)=\frac{2}{3}\cdot1^{\tfrac32}+\frac{3}{2}\cdot1^{\tfrac23}+\frac{4}{5}\cdot1^{\tfrac54} =\frac{2}{3}+\frac{3}{2}+\frac{4}{5} =\frac{20+45+24}{30} =\frac{89}{30}.

Passo 5

Subtraímos F(2)-F(1). Simplificando o que der:
\int_{1}^{2}\Bigl(\sqrt{x}+\tfrac1{\sqrt[3]{x}}+\sqrt[4]{x}\Bigr)\,dx = \Bigl(\frac{4\sqrt{2}}{3}+\frac{3}{2}\,2^{\tfrac23}+\frac{4}{5}\,2^{\tfrac54}\Bigr) -\frac{89}{30}.

Resposta

\boxed{\frac{4\sqrt{2}}{3}+\frac{3}{2}\,2^{\tfrac23}+\frac{4}{5}\,2^{\tfrac54}-\frac{89}{30}}.

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