1. **State the problem:** We are given the integral equation $$\int_1^k \left( \frac{3}{\sqrt{x}} + 4 \right) dx = \frac{95}{4}$$ and need to find the positive constant $k$. 2. **Rewrite the integral:** The integrand is $$\frac{3}{\sqrt{x}} + 4 = 3x^{-\frac{1}{2}} + 4$$. 3. **Find the antiderivative:** $$\int \left(3x^{-\frac{1}{2}} + 4\right) dx = 3 \int x^{-\frac{1}{2}} dx + 4 \int dx = 3 \cdot 2x^{\frac{1}{2}} + 4x + C = 6\sqrt{x} + 4x + C$$ 4. **Evaluate the definite integral:** $$\int_1^k \left( \frac{3}{\sqrt{x}} + 4 \right) dx = \left[6\sqrt{x} + 4x\right]_1^k = (6\sqrt{k} + 4k) - (6\sqrt{1} + 4 \cdot 1) = 6\sqrt{k} + 4k - 6 - 4 = 6\sqrt{k} + 4k - 10$$ 5. **Set equal to given value and solve for $k$:** $$6\sqrt{k} + 4k - 10 = \frac{95}{4}$$ Multiply both sides by 4 to clear denominator: $$4(6\sqrt{k} + 4k - 10) = 95$$ $$24\sqrt{k} + 16k - 40 = 95$$ Add 40 to both sides: $$24\sqrt{k} + 16k = 135$$ 6. **Substitute $t = \sqrt{k}$, so $k = t^2$:** $$24t + 16t^2 = 135$$ Rewrite: $$16t^2 + 24t - 135 = 0$$ 7. **Solve quadratic equation:** Use quadratic formula: $$t = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 16 \cdot (-135)}}{2 \cdot 16} = \frac{-24 \pm \sqrt{576 + 8640}}{32} = \frac{-24 \pm \sqrt{9216}}{32}$$ $$\sqrt{9216} = 96$$ So, $$t = \frac{-24 \pm 96}{32}$$ 8. **Find valid solutions for $t$:** $$t_1 = \frac{-24 + 96}{32} = \frac{72}{32} = \frac{9}{4} = 2.25$$ $$t_2 = \frac{-24 - 96}{32} = \frac{-120}{32} = -\frac{15}{4} = -3.75$$ Since $t = \sqrt{k} \geq 0$, discard negative solution. 9. **Find $k$:** $$k = t^2 = \left(\frac{9}{4}\right)^2 = \frac{81}{16} = 5.0625$$ **Final answer:** $$\boxed{k = \frac{81}{16}}$$