1. **Problem:** Calculate the indefinite integral $\int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \, dx$. 2. **Step 1: Simplify the expression under the square root.** Complete the square for the quadratic inside the root: $$x^2 + 4x + 10 = (x^2 + 4x + 4) + 6 = (x + 2)^2 + 6$$ 3. **Step 2: Use substitution.** Let $u = x + 2$, so $du = dx$ and $x = u - 2$. Rewrite the numerator: $$5x + 3 = 5(u - 2) + 3 = 5u - 10 + 3 = 5u - 7$$ The integral becomes: $$\int \frac{5u - 7}{\sqrt{u^2 + 6}} \, du$$ 4. **Step 3: Split the integral.** $$\int \frac{5u}{\sqrt{u^2 + 6}} \, du - \int \frac{7}{\sqrt{u^2 + 6}} \, du$$ 5. **Step 4: Solve each integral separately.** - For $\int \frac{5u}{\sqrt{u^2 + 6}} \, du$, use substitution $w = u^2 + 6$, so $dw = 2u \, du$. Rewrite: $$\int \frac{5u}{\sqrt{u^2 + 6}} \, du = 5 \int \frac{u}{\sqrt{w}} \, du = 5 \int \frac{u}{\sqrt{w}} \, du$$ Since $dw = 2u \, du$, then $u \, du = \frac{dw}{2}$, so: $$5 \int \frac{u}{\sqrt{w}} \, du = 5 \int \frac{1}{\sqrt{w}} \cdot u \, du = 5 \int \frac{1}{\sqrt{w}} \cdot \frac{dw}{2} = \frac{5}{2} \int w^{-1/2} \, dw$$ Integrate: $$\frac{5}{2} \cdot 2 w^{1/2} = 5 \sqrt{w} = 5 \sqrt{u^2 + 6}$$ - For $\int \frac{7}{\sqrt{u^2 + 6}} \, du$, this is a standard integral: $$\int \frac{1}{\sqrt{u^2 + a^2}} \, du = \ln|u + \sqrt{u^2 + a^2}| + C$$ So: $$7 \int \frac{1}{\sqrt{u^2 + 6}} \, du = 7 \ln|u + \sqrt{u^2 + 6}|$$ 6. **Step 5: Combine results and substitute back $u = x + 2$.** $$\int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \, dx = 5 \sqrt{(x + 2)^2 + 6} - 7 \ln|x + 2 + \sqrt{(x + 2)^2 + 6}| + C$$ **Final answer:** $$\boxed{5 \sqrt{x^2 + 4x + 10} - 7 \ln|x + 2 + \sqrt{x^2 + 4x + 10}| + C}$$ This completes the solution for part (a).