# How do you find the indefinite integral of int (12/x^4+8/x^5) dx?

Mar 3, 2018

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$

#### Explanation:

We can split the integral up, as $\int \left(f \left(x\right) \pm g \left(x\right)\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} \pm \int g \left(x\right) \mathrm{dx} :$

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = \int \left(\frac{12}{x} ^ 4\right) \mathrm{dx} + \int \left(\frac{8}{x} ^ 5\right) \mathrm{dx}$

Let's factor the constants $12$ and $8$ out of their respective integrals:

$\int \left(\frac{12}{x} ^ 4\right) \mathrm{dx} + \int \left(\frac{8}{x} ^ 5\right) \mathrm{dx} = 12 \int \frac{\mathrm{dx}}{x} ^ 4 + 8 \int \frac{\mathrm{dx}}{x} ^ 5$

Let's rewrite our integrals using negative exponents. Recall that $\frac{1}{x} ^ a = {x}^{-} a .$ In other words, bringing a positive exponent from the denominator into the numerator yields the negative form of that exponent.

$12 \int \frac{\mathrm{dx}}{x} ^ 4 + 8 \int \frac{\mathrm{dx}}{x} ^ 5 = 12 \int {x}^{-} 4 \mathrm{dx} + 8 \int {x}^{-} 5 \mathrm{dx}$

Now we can integrate each of these. Recall that $\int {x}^{a} \mathrm{dx} = {x}^{a + 1} / \left(a + 1\right) + C$ where $C$ is the constant of integration.

$12 \int {x}^{-} 4 \mathrm{dx} + 8 \int {x}^{-} 5 \mathrm{dx} = 12 \left({x}^{- 4 + 1} / \left(- 4 + 1\right)\right) + 8 \left({x}^{- 5 + 1} / \left(- 5 + 1\right)\right) + C = \frac{12 {x}^{-} 3}{-} 3 + \frac{8 {x}^{-} 4}{-} 4 + C = - 4 {x}^{-} 3 - 2 {x}^{-} 2 + C$

It may appear confusing that we only have one constant of integration $C$ despite taking two integrals. Yes, each of these integrals did originally produce its own distinct constant of integration; however, we combined them both into a single constant $C .$ Adding two constants yields one constant; therefore, we just have one constant at the end.

We can revert to positive exponents:

$- 4 {x}^{-} 3 - 2 {x}^{-} 2 + C = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$

Thus,

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$