Question

evaluate the indefinite integral.\\(\int \frac{7}{(t + 9)^2} dt\\)\\(\square\\) + c

Explanation:

Step1: Use substitution method

Let $u = t + 9$, then $du = dt$.

Step2: Rewrite integral in terms of u

$$\int \frac{7}{u^2} du = 7\int u^{-2} du$$

Step3: Apply power rule for integration

The power rule is $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n
eq -1$.
$$7 \cdot \frac{u^{-2+1}}{-2+1} = 7 \cdot \frac{u^{-1}}{-1} = -\frac{7}{u}$$

Step4: Substitute back $u = t+9$

$$-\frac{7}{t+9}$$

Answer:

$-\frac{7}{t+9}$