1. **Problem statement:** Find the integral $$\int \frac{1}{(7x+4)^m} \, dx$$ for the cases (a) $$m=2$$ and (b) $$m=1$$. 2. **Formula and rules:** For integrals of the form $$\int (ax+b)^n \, dx$$ where $$n \neq -1$$, the formula is: $$\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$$ For $$n = -1$$, the integral is: $$\int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln|ax+b| + C$$ 3. **Case (a):** $$m=2$$, so the integral becomes: $$\int \frac{1}{(7x+4)^2} \, dx = \int (7x+4)^{-2} \, dx$$ Using the formula for $$n \neq -1$$ with $$n = -2$$ and $$a=7$$: $$\int (7x+4)^{-2} \, dx = \frac{(7x+4)^{-2+1}}{7(-2+1)} + C = \frac{(7x+4)^{-1}}{7(-1)} + C = -\frac{1}{7(7x+4)} + C$$ 4. **Case (b):** $$m=1$$, so the integral becomes: $$\int \frac{1}{7x+4} \, dx$$ Using the formula for $$n = -1$$: $$\int \frac{1}{7x+4} \, dx = \frac{1}{7} \ln|7x+4| + C$$ **Final answers:** - (a) $$\int \frac{1}{(7x+4)^2} \, dx = -\frac{1}{7(7x+4)} + C$$ - (b) $$\int \frac{1}{7x+4} \, dx = \frac{1}{7} \ln|7x+4| + C$$