If \( f(x)=\int\sqrt{\displaystyle\frac{x+8}{x-8}}\,dx;\ f(8)=24\log_{e}2 \), then \( f(10) \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6+32\log_{e}2 \)
Multiplying the numerator and the denominator under the root by \( x+8 \) rationalises the integrand.
\( \displaystyle \sqrt{\frac{x+8}{x-8}}=\frac{x+8}{\sqrt{x^{2}-64}} \)
It splits into two integrals.
\( \displaystyle f(x)=\int\frac{x}{\sqrt{x^{2}-64}}\,dx+8\int\frac{dx}{\sqrt{x^{2}-64}} \)
The first is \( \sqrt{x^{2}-64} \) (put \( t=x^{2}-64 \)); the second is the standard result \( \log_{e}\left|x+\sqrt{x^{2}-64}\right| \). So
\( \displaystyle f(x)=\sqrt{x^{2}-64}+8\log_{e}\left|x+\sqrt{x^{2}-64}\right|+C \)
From \( f(8)=24\log_{e}2 \),
\( \displaystyle 0+8\log_{e}8+C=24\log_{e}2 \)
\( \displaystyle C=0 \)
At x = 10, \( \sqrt{100-64}=6 \), so
\( \displaystyle f(10)=6+8\log_{e}16 \)
\( \displaystyle =6+32\log_{e}2 \)