Question

If \[ \int_{2}^{3} \frac{3\log x} {3\log x+\log(125-75x+15x^2-x^3)} \,dx = k, \] then \[ 4k^2+2k+1= \]

• A. \(9\)
• B. \(3\)
• C. \(25\)
• D. \(\dfrac94\)

Hint:

If an integral contains expressions like \[ f(x) \quad \text{and} \quad f(a+b-x), \] always test the substitution \(x\to a+b-x\). Many definite integrals collapse instantly using symmetry.

Answer 1

The Correct Option is B

Concept: This problem is based on the standard property \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx. \] The complicated logarithmic expression simplifies beautifully after applying this symmetry.

Step 1: Factor the cubic expression. Observe that \[ 125-75x+15x^2-x^3 = (5-x)^3. \] Hence \[ \log(125-75x+15x^2-x^3) = 3\log(5-x). \] Therefore \[ k = \int_2^3 \frac{3\log x} {3\log x+3\log(5-x)} dx. \] Cancelling \(3\), \[ k = \int_2^3 \frac{\log x} {\log x+\log(5-x)} dx. \]

Step 2: Apply the symmetry property. Let \[ I = \int_2^3 \frac{\log x} {\log x+\log(5-x)} dx. \] Replacing \(x\) by \[ 5-x, \] we obtain \[ I = \int_2^3 \frac{\log(5-x)} {\log x+\log(5-x)} dx. \] Adding the two equations, \[ 2I = \int_2^3 1\,dx. \] \[ 2I = 3-2. \] \[ 2I=1. \] \[ I=\frac12. \] Hence \[ k=\frac12. \]

Step 3: Compute the required value. \[ 4k^2+2k+1 = 4\left(\frac12\right)^2 +2\left(\frac12\right) +1. \] \[ = 1+1+1. \] \[ =3. \]

Step 4: Final Answer. \[ \boxed{3} \]