∫₆¹⁶ (log ₑx²)/(log ₑx² + log ₑ(x² - 44x + 484))dx is equal to :

Solution

Maths sir · 4 steps

Spot that x²-44x+484=(x-22)², then use the King's property with a+b=22.

1. Tidy the logs. The quadratic is a perfect square: x²-44x+484=(x-22)². On [6,16] both x and 22-x are positive, so logₑ x² = 2 ln x and logₑ (x-22)² = 2 ln(22-x).
2. Simplified integral. The factor 2 cancels, leaving I=∫₆¹⁶(ln x)/(ln x+ln(22-x)) dx.
3. Apply King's property. Since a+b=6+16=22, the property ∫ₐᵇ f(x) dx=∫ₐᵇ f(a+b-x) dx replaces x by 22-x and gives I=∫₆¹⁶ln(22-x)/(ln(22-x)+ln x) dx.

   2I=∫₆¹⁶(ln x+ln(22-x))/(ln x+ln(22-x)) dx=∫₆¹⁶1 dx=10

4. So. 2I=10, so I=5. The answer is (c).

The idea to keep. When the denominator is f(x)+f(a+b-x) and the numerator is one of them, add the integral to its King's-property twin. The sum is 1, so I=(b-a)/2.

Answer from the NTA final answer key. Solution by PrepWiser.