Question

\(\int \frac{\sqrt{x - 2}}{2x + 4} \, dx =\)

• A. \(\frac{1}{2} \sqrt{x^2 - 2x + 5} + \sinh^{-1}\left( \frac{x - 1}{2} \right) + c\)
• B. \(\sqrt{x - 2} - 2 \tan^{-1}\left( \frac{\sqrt{x - 2}}{2} \right) + c\)
• C. \(\frac{1}{2} \sqrt{x^2 - 2x + 5} - \cosh^{-1}\left( \frac{x - 1}{2} \right) + c\)
• D. \(\frac{1}{2} \sqrt{x - 2} + \tan^{-1}\left( \frac{x - 2}{2} \right) + c\)

Hint:

For integrals with \(\sqrt{x - a}\), try substitution \( u = \sqrt{x - a} \). Rewrite the integrand to use standard forms like \(\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right)\).

Answer 1

The Correct Option is B

Let \( u = \sqrt{x - 2} \), so \( x = u^2 + 2 \), \( dx = 2u \, du \). The denominator becomes: \[ 2x + 4 = 2(u^2 + 2) + 4 = 2u^2 + 8 \] The integral is: \[ \int \frac{u}{2u^2 + 8} \cdot 2u \, du = \int \frac{2u^2}{2(u^2 + 4)} \, du = \int \frac{u^2}{u^2 + 4} \, du \] Rewrite the integrand: \[ \frac{u^2}{u^2 + 4} = \frac{u^2 + 4 - 4}{u^2 + 4} = 1 - \frac{4}{u^2 + 4} \] \[ \int \left( 1 - \frac{4}{u^2 + 4} \right) \, du = u - 4 \int \frac{1}{u^2 + 2^2} \, du \] \[ \int \frac{1}{u^2 + 2^2} \, du = \frac{1}{2} \tan^{-1}\left( \frac{u}{2} \right) \] \[ u - 4 \cdot \frac{1}{2} \tan^{-1}\left( \frac{u}{2} \right) + c = u - 2 \tan^{-1}\left( \frac{u}{2} \right) + c \] Substitute back \( u = \sqrt{x - 2} \): \[ \sqrt{x - 2} - 2 \tan^{-1}\left( \frac{\sqrt{x - 2}}{2} \right) + c \] Option (2) is correct. Options (1), (3), and (4) do not match the derived form.