Question

determine the following indefinite integral.
int(5 s+3)^{2} d s
int(5 s+3)^{2} d s=square

Explanation:

Step1: Expand the integrand

First, we expand \((5s + 3)^2\) using the formula \((a + b)^2=a^2+2ab + b^2\). Here, \(a = 5s\) and \(b=3\), so \((5s + 3)^2=(5s)^2+2\times(5s)\times3+3^2 = 25s^2+30s + 9\).
The integral becomes \(\int(25s^2 + 30s+9)ds\).

Step2: Integrate term - by - term

We use the power rule for integration \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and the constant multiple rule \(\int kf(x)dx=k\int f(x)dx\) (where \(k\) is a constant) and the sum rule \(\int(f(x)+g(x)+h(x))dx=\int f(x)dx+\int g(x)dx+\int h(x)dx\).

• For the first term \(\int25s^2ds\):

Using the constant multiple rule and the power rule, with \(k = 25\) and \(n = 2\), we have \(25\int s^2ds=25\times\frac{s^{2 + 1}}{2+1}=\frac{25}{3}s^3\).

• For the second term \(\int30sds\):

Using the constant multiple rule and the power rule, with \(k = 30\) and \(n = 1\), we have \(30\int sds=30\times\frac{s^{1+1}}{1 + 1}=30\times\frac{s^2}{2}=15s^2\).

• For the third term \(\int9ds\):

Using the rule \(\int kdx=kx + C\) (where \(k\) is a constant), we have \(\int9ds=9s\).

Step3: Combine the results and add the constant of integration

Combining the three results and adding the constant of integration \(C\) (since it's an indefinite integral), we get \(\frac{25}{3}s^3+15s^2 + 9s+C\).

We can also solve this integral using substitution. Let \(u=5s + 3\), then \(du = 5ds\), or \(ds=\frac{1}{5}du\).

The integral \(\int(5s + 3)^2ds=\int u^2\times\frac{1}{5}du=\frac{1}{5}\times\frac{u^{3}}{3}+C=\frac{(5s + 3)^{3}}{15}+C\).

Let's expand \(\frac{(5s + 3)^{3}}{15}\):
\((5s+3)^{3}=(5s)^{3}+3\times(5s)^{2}\times3+3\times(5s)\times3^{2}+3^{3}=125s^{3}+225s^{2}+135s + 27\)
\(\frac{125s^{3}+225s^{2}+135s + 27}{15}=\frac{125}{15}s^{3}+\frac{225}{15}s^{2}+\frac{135}{15}s+\frac{27}{15}=\frac{25}{3}s^{3}+15s^{2}+9s+\frac{9}{5}\). The \(\frac{9}{5}\) is absorbed into the constant \(C\) when we write the general form of the antiderivative. So both methods give equivalent results (up to the constant of integration).

Answer:

\(\frac{(5s + 3)^{3}}{15}+C\) (or \(\frac{25}{3}s^3+15s^2 + 9s+C\))