# How do you find the integral of 20+ (4s^4)/sqrts ds?

Jun 3, 2016

$20 s + \frac{8}{9} {s}^{\frac{9}{2}} + C$

#### Explanation:

We have

$\int \left(20 + \frac{4 {s}^{4}}{\sqrt{s}}\right) \mathrm{ds}$

Split this up since integrals can be separated through addition:

$= \int 20 \mathrm{ds} + \int \frac{4 {s}^{4}}{\sqrt{s}} \mathrm{ds}$

The second integrand can be simplified as follows:

$\frac{4 {s}^{4}}{\sqrt{s}} = \frac{4 {s}^{4}}{s} ^ \left(\frac{1}{2}\right) = 4 {s}^{4 - \frac{1}{2}} = 4 {s}^{\frac{7}{2}}$

So we have the integral

$\int 20 \mathrm{ds} + 4 \int {s}^{\frac{7}{2}} \mathrm{ds}$

The first integral is just $20 s + C$, and find the next integral using the rule $\int {s}^{n} \mathrm{ds} = {s}^{n + 1} / \left(n + 1\right) + C$.

$= 20 s + 4 \left({s}^{\frac{7}{2} + 1} / \left(\frac{7}{2} + 1\right)\right) + C$

$= 20 s + 4 \left(\frac{2}{9}\right) {s}^{\frac{9}{2}} + C$

$= 20 s + \frac{8}{9} {s}^{\frac{9}{2}} + C$