\( \int_{4}^{10}\displaystyle\frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx \) is equal to:
(Consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
The second logarithm is of a perfect square:
\( \displaystyle 196-28x+x^2=(14-x)^2 \)
So
\( \displaystyle I=\int_{4}^{10}\displaystyle\frac{\log(x^2)}{\log(x^2)+\log\left((14-x)^2\right)}\,dx \)
By the property \( \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx \) with a + b = 14, replacing x by 14 − x changes \( x^2 \) into \( (14-x)^2 \) and \( (14-x)^2 \) into \( x^2 \):
\( \displaystyle I=\int_{4}^{10}\displaystyle\frac{\log\left((14-x)^2\right)}{\log\left((14-x)^2\right)+\log(x^2)}\,dx \)
Adding the two forms of I, the numerators add up to the denominator:
\( \displaystyle 2I=\int_{4}^{10}\displaystyle 1\,dx \)
\( \displaystyle 2I=10-4=6 \)
\( \displaystyle I=3 \)
Tip: When the integrand is \( \frac{g(x)}{g(x)+g(a+b-x)} \) over [a, b], the answer is always \( \frac{b-a}{2} \).