exercise 1
\[\begin{eqnarray}
\int \frac{1}{\sqrt[]{x}}dx
\newline \newline
\int \frac{1}{\sqrt[]{x}}dx=\int \frac{1}{x^{\frac{1}{2}}}dx=\int x^{-\frac{1}{2}}dx
\newline \newline
\Rightarrow \frac{1}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}+C
\newline \newline
2 \times x^{\frac{1}{2}}+C
\newline \newline
2\sqrt[]{x} +C
\end{eqnarray}
\]
exercise 2
\[
\begin{align}
\int x^{2} \times \sqrt[3]{x} dx
\\ \\
\because x^{2} \times \sqrt[3]{x} \Rightarrow x^{2+\frac{1}{3} }=x^{\frac{7}{3} }
\\ \\
\therefore \int x^{\frac{7}{3} }dx= \frac{1}{\frac{7}{3}+1} x^{\frac{7}{3} +1 } + C
\\ \\
\Rightarrow \frac{3}{10} x^{\frac{10}{3} } + C
\end{align}
\]
exercise 3
\[\begin{align}
\int(\sqrt{x}+1)(x-\sqrt{x}) d x=?
\\ \\
\Rightarrow \int\left(x^{\frac{1}{2}}+1\right)\left(x-x^{-\frac{1}{2}}\right) d x
\\ \\
= \int(x^{\frac{3}{2}}-1+x-x^{-\frac{1}{2}} )dx
\\ \\
\int x^{\frac{3}{2}} d x=\frac{1}{\frac{3}{2}+1} x^{\frac{3}{2}+1}=\frac{2}{5} x^{\frac{5}{2}}
\\ \\
\int x d x=\frac{1}{1+1} x^{1+1}=\frac{1}{2} x^{2}
\\ \\
\int x^{-\frac{1}{2}} d x=\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}=2 x^{\frac{1}{2}}
\\ \\
\int 1 d x=x
\\ \\
\Rightarrow \frac{2}{5} x^{\frac{5}{2}}-x+\frac{1}{2} x^{2}-2 x^{\frac{1}{2}}+C
\end{align}
\]