If f(x)= √((x+8)/(x-8)) dx; f(8)=24_e 2, then f(10) is equal to: Option 1: 6+32_e 2 Option 2: 6+^(-1)4/5 Option 3: -5+^(-1)4/5 Option 4: 6+1/8_e 4/5

Answer: 6+32_e 2. Solution: Rationalise: √((x+8)/(x-8)) = (x+8)/(√(x^2-64)). So f(x) = √(x^2-64) + 8_e, x+√(x^2-64), + C. At x=8: 0 + 8_e 8 = 24_e 2, so C=0. Then f(10) = 6 + 8_e 16 = 6 + 32_e 2.

Calculus

Integrals

Hard

core

If $f(x)=\int \sqrt{\dfrac{x+8}{x-8}}\,dx; f(8)=24\log_e 2$ , then $f(10)$ is equal to:

$6+32\log_e 2$

$6+\tan^{-1}\dfrac{4}{5}$

$-5+\sin^{-1}\dfrac{4}{5}$

$6+\dfrac{1}{8}\log_e\left|\dfrac{4}{5}\right|$

✅ Correct Option: 1

Rationalise: $\sqrt{\dfrac{x+8}{x-8}} = \dfrac{x+8}{\sqrt{x^2-64}}$ .

So $f(x) = \sqrt{x^2-64} + 8\log_e|x+\sqrt{x^2-64}| + C$ .

At $x=8$ : $0 + 8\log_e 8 = 24\log_e 2$ , so $C=0$ .

Then $f(10) = 6 + 8\log_e 16 = 6 + 32\log_e 2$ .

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If f(x)=∫x+8x−8 dx;f(8)=24log⁡e2f(x)=\int \sqrt{\dfrac{x+8}{x-8}}\,dx; f(8)=24\log_e 2f(x)=∫x−8x+8​​dx;f(8)=24loge​2, then f(10)f(10)f(10) is equal to:

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If $f(x)=\int \sqrt{\dfrac{x+8}{x-8}}\,dx; f(8)=24\log_e 2$ , then $f(10)$ is equal to: