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: $$5x + 3 = 5(u - 2) + 3 = 5u - 10 + 3 = 5u - 7$$ 4. **Step 3: Rewrite the integral in terms of $$u$$:** $$\int \frac{5u - 7}{\sqrt{u^2 + 6}} \, du = 5 \int \frac{u}{\sqrt{u^2 + 6}} \, du - 7 \int \frac{1}{\sqrt{u^2 + 6}} \, du$$ 5. **Step 4: Solve each integral separately.** - For $$\int \frac{u}{\sqrt{u^2 + 6}} \, du$$, use substitution: Let $$w = u^2 + 6$$, then $$dw = 2u \, du$$, so $$u \, du = \frac{dw}{2}$$. Thus, $$\int \frac{u}{\sqrt{u^2 + 6}} \, du = \int \frac{1}{\sqrt{w}} \cdot \frac{dw}{2} = \frac{1}{2} \int w^{-1/2} dw = \frac{1}{2} \cdot 2 w^{1/2} + C = \sqrt{u^2 + 6} + C$$ - For $$\int \frac{1}{\sqrt{u^2 + 6}} \, du$$, this is a standard integral: $$\int \frac{1}{\sqrt{u^2 + a^2}} \, du = \ln|u + \sqrt{u^2 + a^2}| + C$$ with $$a^2 = 6$$. 6. **Step 5: Combine results:** $$5 \sqrt{u^2 + 6} - 7 \ln|u + \sqrt{u^2 + 6}| + C$$ 7. **Step 6: Substitute back $$u = x + 2$$:** $$\boxed{5 \sqrt{(x+2)^2 + 6} - 7 \ln|x + 2 + \sqrt{(x+2)^2 + 6}| + C}$$ This is the final answer for the first integral. --- **Note:** The user asked multiple integrals but per instructions, only the first problem is solved.