Question

If $$f(x) = \begin{cases} \frac{6x^2 + 1}{4x^3 + 2x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \leq x < 2 \end{cases}$$ then $$\int_{0}^{2} f(x) \,dx = ?$$

• A. $\frac{1}{2} \log 3 + \frac{10}{3}$
• B. $\frac{1}{2} \log 3 - \frac{10}{3}$
• C. $\frac{1}{2} \log 3 + \frac{13}{3}$
• D. $\frac{1}{2} \log 3 + \frac{20}{3}$

Hint:

For integrals involving rational functions, check if the numerator is a derivative of the denominator. For polynomial functions, use standard integration rules.

Answer 1

The Correct Option is A

Step 1: Splitting the integral
We need to compute: $$I = \int_{0}^{2} f(x) \,dx.$$ Since $f(x)$ is defined in two parts, we split the integral: $$I = \int_{0}^{1} \frac{6x^2 + 1}{4x^3 + 2x + 3} \,dx + \int_{1}^{2} (x^2 + 1) \,dx.$$ Step 2: Evaluating the first integral
Consider: $$I_1 = \int_{0}^{1} \frac{6x^2 + 1}{4x^3 + 2x + 3} \,dx.$$ Observing the denominator: $$4x^3 + 2x + 3.$$ Differentiating: $$\frac{d}{dx} (4x^3 + 2x + 3) = 12x^2 + 2.$$ Rewriting the numerator: $$6x^2 + 1 = \frac{1}{2} (12x^2 + 2).$$ Thus, rewriting the integral: $$I_1 = \int_{0}^{1} \frac{\frac{1}{2} (12x^2 + 2)}{4x^3 + 2x + 3} \,dx.$$ $$= \frac{1}{2} \int_{0}^{1} \frac{d(4x^3 + 2x + 3)}{4x^3 + 2x + 3}.$$ $$= \frac{1}{2} \log |4x^3 + 2x + 3| \Big|_{0}^{1}.$$ Evaluating at limits: $$I_1 = \frac{1}{2} \log \frac{|4(1)^3 + 2(1) + 3|}{|4(0)^3 + 2(0) + 3|}.$$ $$= \frac{1}{2} \log \frac{4 + 2 + 3}{3}.$$ $$= \frac{1}{2} \log \frac{9}{3} = \frac{1}{2} \log 3.$$ Step 3: Evaluating the second integral
$$I_2 = \int_{1}^{2} (x^2 + 1) \,dx.$$ $$= \left[ \frac{x^3}{3} + x \right]_{1}^{2}.$$ $$= \left( \frac{2^3}{3} + 2 \right) - \left( \frac{1^3}{3} + 1 \right).$$ $$= \left( \frac{8}{3} + 2 \right) - \left( \frac{1}{3} + 1 \right).$$ $$= \left( \frac{8}{3} + \frac{6}{3} \right) - \left( \frac{1}{3} + \frac{3}{3} \right).$$ $$= \frac{14}{3} - \frac{4}{3} = \frac{10}{3}.$$ Step 4: Conclusion
Adding both integrals: $$I = \frac{1}{2} \log 3 + \frac{10}{3}.$$ Thus, the correct answer is: $$\frac{1}{2} \log 3 + \frac{10}{3}.$$