Let's evaluate the integral [tex]\( I = \int \left( 5x^{\frac{1}{4}} - \frac{1}{x^3} + 15x^4 - 9 \right) \, dx \)[/tex].
To solve this, we will integrate each term of the integrand separately.
1. Integrate [tex]\( 5x^{\frac{1}{4}} \)[/tex]:
[tex]\[
\int 5x^{\frac{1}{4}} \, dx
\][/tex]
The power rule of integration states that [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1}\)[/tex], provided [tex]\( n \neq -1\)[/tex]. Here, [tex]\( n = \frac{1}{4} \)[/tex]:
[tex]\[
\int 5x^{\frac{1}{4}} \, dx = 5 \cdot \frac{x^{\frac{1}{4} + 1}}{\frac{1}{4} + 1} = 5 \cdot \frac{x^{\frac{5}{4}}}{\frac{5}{4}} = 5 \cdot \frac{4}{5} x^{\frac{5}{4}} = 4x^{\frac{5}{4}}
\][/tex]
2. Integrate [tex]\( -\frac{1}{x^3} \)[/tex] (or [tex]\( -x^{-3} \)[/tex]):
[tex]\[
\int -x^{-3} \, dx
\][/tex]
Again, using the power rule of integration with [tex]\( n = -3 \)[/tex]:
[tex]\[
\int -x^{-3} \, dx = - \frac{x^{-3+1}}{-3+1} = - \frac{x^{-2}}{-2} = \frac{x^{-2}}{2} = \frac{1}{2} x^{-2} = \frac{1}{2} \cdot \frac{1}{x^2} = \frac{1}{2x^2}
\][/tex]
3. Integrate [tex]\( 15x^4 \)[/tex]:
[tex]\[
\int 15x^4 \, dx
\][/tex]
Using the power rule with [tex]\( n = 4 \)[/tex]:
[tex]\[
\int 15x^4 \, dx = 15 \cdot \frac{x^{4+1}}{4+1} = 15 \cdot \frac{x^5}{5} = 3x^5
\][/tex]
4. Integrate [tex]\( -9 \)[/tex]:
[tex]\[
\int -9 \, dx
\][/tex]
Since the integral of a constant [tex]\( C \)[/tex] is [tex]\( Cx \)[/tex]:
[tex]\[
\int -9 \, dx = -9x
\][/tex]
Putting all these results together, we get:
[tex]\[
I = 4x^{\frac{5}{4}} + \frac{1}{2x^2} + 3x^5 - 9x + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
Thus, the evaluated integral is:
[tex]\[
I = 3x^5 - 9x + 4x^{\frac{5}{4}} + \frac{1}{2x^2} + C
\][/tex]
