Question

determine the following indefinite integral.
int(7s + 4)^{2}ds
int(7s + 4)^{2}ds=square

Explanation:

Step1: Expand the integrand

Use the formula $(a + b)^2=a^{2}+2ab + b^{2}$, so $(7s + 4)^{2}=(7s)^{2}+2\times7s\times4 + 4^{2}=49s^{2}+56s + 16$. Then the integral becomes $\int(49s^{2}+56s + 16)ds$.

Step2: Integrate term - by - term

According to the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$).
For $\int49s^{2}ds=49\times\frac{s^{3}}{3}=\frac{49s^{3}}{3}$, for $\int56sds=56\times\frac{s^{2}}{2}=28s^{2}$, and for $\int16ds=16s$.

Step3: Combine the results

$\int(49s^{2}+56s + 16)ds=\frac{49s^{3}}{3}+28s^{2}+16s + C$.

Answer:

$\frac{49s^{3}}{3}+28s^{2}+16s + C$