Question

\( \int \frac{1}{\sqrt{9+8x-x^2}}\,dx = \varphi(x) + c \), then \( \varphi(x) = \)

• A. \( \frac{1}{5}\sin^{-1}\left(\frac{x-4}{5}\right) \)
• B. \( \sin^{-1}\left(\frac{x-4}{5}\right) \)
• C. \( \frac{1}{10}\log\left|\frac{4-x}{4+x}\right| \)
• D. \( \log\left|\frac{4-x}{4+x}\right| \)

Hint:

For integrals involving \( \sqrt{a^2 - (x-h)^2} \), always complete the square and use inverse trigonometric substitution.

Answer 1

The Correct Option is B

Step 1: Simplify the expression inside square root.
\[ 9 + 8x - x^2 = -(x^2 - 8x - 9). \]

Step 2: Complete the square.
\[ x^2 - 8x = (x-4)^2 - 16. \]
So,
\[ x^2 - 8x - 9 = (x-4)^2 - 25. \]
Thus,
\[ 9 + 8x - x^2 = 25 - (x-4)^2. \]

Step 3: Rewrite the integral.
\[ \int \frac{1}{\sqrt{25-(x-4)^2}}\,dx. \]

Step 4: Use standard integral formula.
\[ \int \frac{1}{\sqrt{a^2 - u^2}}\,du = \sin^{-1}\left(\frac{u}{a}\right) + C. \]
Here \( u = x-4 \), \( a = 5 \).

Step 5: Apply substitution.
\[ \int \frac{1}{\sqrt{25-(x-4)^2}}\,dx = \sin^{-1}\left(\frac{x-4}{5}\right) + C. \]

Step 6: Identify \( \varphi(x) \).
Thus,
\[ \varphi(x) = \sin^{-1}\left(\frac{x-4}{5}\right). \]

Step 7: Final conclusion.
Hence, the correct answer is option (B).
Final Answer:
\[ \boxed{\sin^{-1}\left(\frac{x-4}{5}\right)}. \]