How do you find the indefinite integral of #int (-2x^-3+10x^-5) dx#?

Question

Samantha Adkison · Accepted Answer

#1/x^2-5/(2x^4)+c# integrate each term using the #color(blue)"power rule for integration"# #color(red)(|bar(ul(color(white)(a/a)color(black)(int(ax^n)dx=a/(n+1)x^(n+1))color(white)(a/a)|)))# #rArr(-2)/(-2)x^-2+10/(-4)x^-4+c# #rArrint(-2x^-3+10x^-5)dx=1/x^2-5/(2x^4)+c#

Emily Ammerman · Answer

undefined To find the indefinite integral of $\int (-2x^{-3} + 10x^{-5}) \, dx$, you can use the power rule for integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$ and $C$ is the constant of integration.

So applying the power rule:

$$\int (-2x^{-3} + 10x^{-5}) \, dx = -2 \int x^{-3} \, dx + 10 \int x^{-5} \, dx$$

For $ \int x^{-3} \, dx$, $n = -3$, so:

$$ \int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C = -\frac{1}{2}x^{-2} + C$$

For $ \int x^{-5} \, dx$, $n = -5$, so:

$$ \int x^{-5} \, dx = \frac{x^{-5+1}}{-5+1} + C = -\frac{1}{4}x^{-4} + C$$

Therefore, combining the results:

$$\int (-2x^{-3} + 10x^{-5}) \, dx = -2 \left(-\frac{1}{2}x^{-2}\right) + 10 \left(-\frac{1}{4}x^{-4}\right) + C$$
$$= x^{-2} - \frac{5}{2}x^{-4} + C$$
$$= \frac{1}{x^2} - \frac{5}{2x^4} + C$$

So, the indefinite integral of $\int (-2x^{-3} + 10x^{-5}) \, dx$ is $\frac{1}{x^2} - \frac{5}{2x^4} + C$, where $C$ is the constant of integration.