Question

find the indefinite integral and check the result by differentiating.

\\( \int(9 x+9)^{5}(9) d x \\)

\\( \int(9 x+9)^{5}(9) d x= \\)

Explanation:

Step1: Expand the integrand

$$\begin{align*} \int(9x + 9)^5(9)dx&=\int(9x + 9)^5\cdot9dx \end{align*}$$

Let \(u = 9x+9\), then \(du=9dx\).

Step2: Integrate using the power rule

Using the power rule \(\int u^n du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 5\) and \(u=9x + 9\), \(du = 9dx\)

$$\begin{align*} \int(9x + 9)^5\cdot9dx&=\int u^5du\\ &=\frac{u^{6}}{6}+C \end{align*}$$

Step3: Substitute back \(u = 9x+9\)

Substitute \(u = 9x + 9\) into \(\frac{u^{6}}{6}+C\)
$$\frac{(9x + 9)^{6}}{6}+C$$

Step4: Check by differentiating

Differentiate \(y=\frac{(9x + 9)^{6}}{6}+C\) using the chain rule. If \(y=\frac{(9x + 9)^{6}}{6}+C\), let \(u = 9x+9\), \(y=\frac{u^{6}}{6}+C\)
\(y^\prime=\frac{6u^{5}}{6}\cdot9=(9x + 9)^5\cdot9\) (by the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), \(\frac{dy}{du}=u^{5}\), \(\frac{du}{dx}=9\))

Answer:

\(\frac{(9x + 9)^{6}}{6}+C\)