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Question: \(\int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (4 + x)}}{{\ln (4 + x) + \ln (9 - x)}}dx} \) is equal to...

∫2−ln⁡33+ln⁡3ln⁡(4+x)ln⁡(4+x)+ln⁡(9−x)dx\int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (4 + x)}}{{\ln (4 + x) + \ln (9 - x)}}dx} is equal to:
1. Cannot be evaluated
2. Is equal to 52\dfrac{5}{2}
3. Is equal to 1+2ln⁡31 + 2\ln 3
4. Is equal to 12+ln⁡3\dfrac{1}{2} + \ln 3

Explanation

Solution

We have to integrate the given complex function. It is a definite function so we will get a constant answer and we will not add any constant of integration at the end. We will use formulas of integration to solve it. It is going to be very difficult to solve it directly so we will use a different formula to solve this equation.

Complete step by step solution:
According to the question we have to integrate ∫2−ln⁡33+ln⁡3ln⁡(4+x)ln⁡(4+x)+ln⁡(9−x)dx\int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (4 + x)}}{{\ln (4 + x) + \ln (9 - x)}}dx}
So, we will use the integration formula ∫xaf(x)=∫baf(a+b−x)\int_x^a {f(x) = } \int_b^a {f(a + b - x)} , this is also called as King’s rule.
This formula represents that we can add both the limits with the variable of the function in the integration and yet, the value will not change. It helps to simplify the equation and solve the equation easily.
So, now let us say I=∫2−ln⁡33+ln⁡3ln⁡(4+x)ln⁡(4+x)+ln⁡(9−x)dxI = \int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (4 + x)}}{{\ln (4 + x) + \ln (9 - x)}}dx} (Consider it as equation 1)
Now, we will use king’s rule in II
So we will get
⇒I=∫2−ln⁡33+ln⁡3ln⁡(4+3+ln⁡3+2−ln⁡3−x)ln⁡(4+3+ln⁡3+2−ln⁡3−x)+ln⁡(9−3−ln⁡3−2+ln⁡3+x)dx\Rightarrow I = \int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (4 + 3 + \ln 3 + 2 - \ln 3 - x)}}{{\ln (4 + 3 + \ln 3 + 2 - \ln 3 - x) + \ln (9 - 3 - \ln 3 - 2 + \ln 3 + x)}}dx}
It can be solved as
⇒I=∫2−ln⁡33+ln⁡3ln⁡(9−x)ln⁡(9−x)+ln⁡(4+x)dx\Rightarrow I = \int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (9 - x)}}{{\ln (9 - x) + \ln (4 + x)}}dx} (Consider it as equation 2)
Now, we will add equation one and equation two, so we will get
⇒2I=∫2−ln⁡33+ln⁡3ln⁡(9−x)+ln⁡(4+x)ln⁡(9−x)+ln⁡(4+x)dx\Rightarrow 2I = \int_{2 - \ln 3}^{3 + \ln 3} {\dfrac{{\ln (9 - x) + \ln (4 + x)}}{{\ln (9 - x) + \ln (4 + x)}}dx}
⇒2I=∫2−ln⁡33+ln⁡31dx\Rightarrow 2I = \int_{2 - \ln 3}^{3 + \ln 3} {1dx} (As numerator and denominator are equal)
⇒2I=3+ln⁡3−2+ln⁡3\Rightarrow 2I = 3 + \ln 3 - 2 + \ln 3 (Integration of one is x)
⇒I=12(1+2ln⁡3)\Rightarrow I = \dfrac{1}{2}(1 + 2\ln 3)
⇒I=12+ln⁡3\Rightarrow I = \dfrac{1}{2} + \ln 3

Hence, option four is our answer.

Note:
We have used king’s rule to solve it because we can’t solve the equation directly easily. It helps to simplify the equation and solve the equation easily. We basically add the new equation we get after using king’s rule and the previous equation to solve the questions in which we use king’s rule. We need to know the integration rule to solve these types of questions.