Question:

Step 1: Understand the problem using differentiation. Since we're given that $\int x^{26}(x-1)^{17}(5x-3)\, dx = \dfrac{x^{27}(x-1)^{18}}{k} + C$, we can verify this by differentiating the right-hand side and checking if it equals the integrand on the left-hand side. Step 2: Differentiate the right-hand side using the product rule. We differentiate $\dfrac{x^{27}(x-1)^{18}}{k}$ with respect to $x$: $$\dfrac{d}{dx}\left[\dfrac{x^{27}(x-1)^{18}}{k}\right] = \dfrac{1}{k}\dfrac{d}{dx}\left[x^{27}(x-1)^{18}\right]$$ Using the product rule: $$= \dfrac{1}{k}\left[27x^{26}(x-1)^{18} + x^{27} \cdot 18(x-1)^{17}\right]$$ Step 3: Factor out the common terms. We can factor out $x^{26}(x-1)^{17}$ from both terms: $$= \dfrac{1}{k} \cdot x^{26}(x-1)^{17}\left[27(x-1) + 18x\right]$$ Step 4: Simplify the expression in brackets. Expanding the bracket: $$27(x-1) + 18x = 27x - 27 + 18x = 45x - 27$$ We can factor out 9: $$45x - 27 = 9(5x - 3)$$ Step 5: Substitute back and simplify. $$\dfrac{d}{dx}\left[\dfrac{x^{27}(x-1)^{18}}{k}\right] = \dfrac{1}{k} \cdot x^{26}(x-1)^{17} \cdot 9(5x-3)$$ $$= \dfrac{9x^{26}(x-1)^{17}(5x-3)}{k}$$ Step 6: Match coefficients with the integrand. For this derivative to equal the original integrand $x^{26}(x-1)^{17}(5x-3)$, we need: $$\dfrac{9x^{26}(x-1)^{17}(5x-3)}{k} = x^{26}(x-1)^{17}(5x-3)$$ This requires: $$\dfrac{9}{k} = 1$$ Therefore: $$k = 9$$ The value of $k$ is **9**, which corresponds to **Option 3: (c)**.