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MHT CET · Maths · Indefinite Integration

\(\int \sqrt{x^2-6 x-16} \mathrm{~d} x\) equals.

  1. A \(\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}+\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+\mathrm{c}\) where \(c\) is the constant of integration
  2. B \(\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}-\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c\) where \(c\) is the constant of integration
  3. C \(\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}+\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+\mathrm{c}\) where \(c\) is the constant of integration
  4. D \(\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}-\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c\), where \(c\) is the constant of integration
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}-\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c\) where \(c\) is the constant of integration

Step-by-step Solution

Detailed explanation

\(\int \sqrt{x^2-6 x-16} \mathrm{~d} x = \int \sqrt{(x-3)^2 - 5^2} \mathrm{~d} x\) \(= \frac{x-3}{2} \sqrt{(x-3)^2 - 5^2} - \frac{5^2}{2} \log \left|x-3+\sqrt{(x-3)^2 - 5^2}\right|+\mathrm{c}\)