Example 3: Using Indefinite Integral Properties and Formulas (A) $$ \int \frac{5}{x^{4}} d x= $$ (1) $$ \int 4 \sqrt[3]{t^{2}} d t= $$ (C) $$ \int \frac{x^{4}-2}{x^{2}} d x= $$ (D) $$ \int\left(\frac{2}{\sqrt[3]{x}}-6 \sqrt{x}\right) d x= $$

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**(A)**  
We start with  
$$
\int \frac{5}{x^4}\,dx = 5\int x^{-4}\,dx.
$$  
Using the power rule,  
$$
\int x^{n}\,dx = \frac{x^{n+1}}{n+1}+C,\quad n\neq -1,
$$  
with $$n=-4$$ we have  
$$
\int x^{-4}\,dx = \frac{x^{-3}}{-3}+C.
$$  
Thus,  
$$
5\int x^{-4}\,dx = 5\left(\frac{x^{-3}}{-3}\right)+C = -\frac{5}{3}x^{-3}+C.
$$

**(1)**  
Express the cube root as an exponent:  
$$
\sqrt[3]{t^2} = t^{\frac{2}{3}}.
$$  
Then,  
$$
\int 4 t^{\frac{2}{3}}\,dt = 4\int t^{\frac{2}{3}}\,dt.
$$  
Using the power rule with $$n=\frac{2}{3}$$:  
$$
\int t^{\frac{2}{3}}\,dt = \frac{t^{\frac{5}{3}}}{\frac{5}{3}}+C = \frac{3}{5}t^{\frac{5}{3}}+C.
$$  
Hence,  
$$
4\int t^{\frac{2}{3}}\,dt = 4\left(\frac{3}{5}t^{\frac{5}{3}}\right)+C = \frac{12}{5}t^{\frac{5}{3}}+C.
$$

**(C)**  
Write the integrand by dividing each term:  
$$
\frac{x^4-2}{x^2} = x^{4-2} - \frac{2}{x^2} = x^2 - 2x^{-2}.
$$  
Separate the integrals:  
$$
\int (x^2-2x^{-2})\,dx = \int x^2\,dx - 2\int x^{-2}\,dx.
$$  
Integrate each term using the power rule:

For $$\int x^2\,dx$$:  
$$
\int x^2\,dx = \frac{x^3}{3}+C.
$$

For $$\int x^{-2}\,dx$$ with $$n=-2$$:  
$$
\int x^{-2}\,dx = \frac{x^{-1}}{-1}+C = -x^{-1}+C.
$$
Multiplying by $$-2$$:  
$$
-2\int x^{-2}\,dx = -2(-x^{-1}) = 2x^{-1}.
$$

Thus, the result is:  
$$
\frac{x^3}{3}+2x^{-1}+C.
$$

**(D)**  
Express the radicals as exponents:  
$$
\frac{2}{\sqrt[3]{x}} = 2x^{-\frac{1}{3}},\quad \sqrt{x} = x^{\frac{1}{2}}.
$$  
Then,  
$$
\int\left(\frac{2}{\sqrt[3]{x}}-6\sqrt{x}\right)dx = 2\int x^{-\frac{1}{3}}\,dx - 6\int x^{\frac{1}{2}}\,dx.
$$

For $$\int x^{-\frac{1}{3}}\,dx$$, using the power rule with $$n=-\frac{1}{3}$$:  
$$
\int x^{-\frac{1}{3}}\,dx = \frac{x^{\frac{2}{3}}}{\frac{2}{3}}+C = \frac{3}{2}x^{\frac{2}{3}}+C.
$$
Multiplying by $$2$$:  
$$
2\int x^{-\frac{1}{3}}\,dx = 2\left(\frac{3}{2}x^{\frac{2}{3}}\right)+C = 3x^{\frac{2}{3}}+C.
$$

For $$\int x^{\frac{1}{2}}\,dx$$ with $$n=\frac{1}{2}$$:  
$$
\int x^{\frac{1}{2}}\,dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C = \frac{2}{3}x^{\frac{3}{2}}+C.
$$
Multiplying by $$-6$$:  
$$
-6\int x^{\frac{1}{2}}\,dx = -6\left(\frac{2}{3}x^{\frac{3}{2}}\right)+C = -4x^{\frac{3}{2}}+C.
$$

Combining the results:  
$$
3x^{\frac{2}{3}}-4x^{\frac{3}{2}}+C.
$$
**(A)** $$ \int \frac{5}{x^4}\,dx = -\frac{5}{3x^3} + C $$ **(1)** $$ \int 4 \sqrt[3]{t^2}\,dt = \frac{12}{5}t^{\frac{5}{3}} + C $$ **(C)** $$ \int \frac{x^4 - 2}{x^2}\,dx = \frac{x^3}{3} + \frac{2}{x} + C $$ **(D)** $$ \int \left(\frac{2}{\sqrt[3]{x}} - 6\sqrt{x}\right)\,dx = 3x^{\frac{2}{3}} - 4x^{\frac{3}{2}} + C $$

Example 3: Using Indefinite Integral Properties and Formulas
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