Question

\(\int \frac{x}{\sqrt{x^2 - 2x + 5}} \, dx =\)

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

Hint:

For integrals like \(\int \frac{x}{\sqrt{x^2 + ax + b}} \, dx\), complete the square and use substitution or hyperbolic functions (\( \int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}\left( \frac{x}{a} \right) \)).

Answer 1

The Correct Option is A

Complete the square: \[ x^2 - 2x + 5 = (x - 1)^2 - 1 + 5 = (x - 1)^2 + 4 \] \[ \sqrt{x^2 - 2x + 5} = \sqrt{(x - 1)^2 + 2^2} \] Let \( u = x - 1 \), so \( x = u + 1 \), \( dx = du \): \[ \int \frac{u + 1}{\sqrt{u^2 + 4}} \, du = \int \frac{u}{\sqrt{u^2 + 4}} \, du + \int \frac{1}{\sqrt{u^2 + 4}} \, du \] First integral: Let \( v = u^2 + 4 \), \( dv = 2u \, du \), \( u \, du = \frac{dv}{2} \): \[ \int \frac{u}{\sqrt{u^2 + 4}} \cdot \frac{du}{u} = \int \frac{1}{\sqrt{v}} \cdot \frac{dv}{2} = \frac{1}{2} \int v^{-1/2} \, dv = \sqrt{v} = \sqrt{u^2 + 4} = \sqrt{x^2 - 2x + 5} \] Second integral: \[ \int \frac{du}{\sqrt{u^2 + 2^2}} = \sinh^{-1}\left( \frac{u}{2} \right) + c = \sinh^{-1}\left( \frac{x - 1}{2} \right) + c \] Alternatively, use hyperbolic substitution: Let \( u = 2 \sinh \theta \), \( du = 2 \cosh \theta \, d\theta \), \( \sqrt{u^2 + 4} = 2 \cosh \theta \): \[ \int \frac{u + 1}{\sqrt{u^2 + 4}} \, du = \int \frac{2 \sinh \theta + 1}{2 \cosh \theta} \cdot 2 \cosh \theta \, d\theta = \int (2 \sinh \theta + 1) \, d\theta \] \[ = 2 \cosh \theta + \theta + c = 2 \sqrt{\frac{u^2 + 4}{4}} + \sinh^{-1}\left( \frac{u}{2} \right) + c = \sqrt{u^2 + 4} + \sinh^{-1}\left( \frac{x - 1}{2} \right) + c \] \[ = \sqrt{x^2 - 2x + 5} + \sinh^{-1}\left( \frac{x - 1}{2} \right) + c \] Option (1) is correct. Options (2), (3), and (4) do not match.