Question:

Evaluate the integral: \[ \int \frac{dx}{(x+1)\sqrt{x^2 + 4}} \]

Show Hint

For integrals involving square roots of quadratic expressions, try using hyperbolic substitutions like \( x = a \sinh t \) or trigonometric substitutions.
Updated On: Mar 17, 2025
  • \( \frac{1}{2} \frac{x+1}{\sqrt{x+2}} + C \)
  • \( \log \left| \frac{x+2}{x+1} \right| + C \)
  • \( \frac{1}{\sqrt{5}} \sinh^{-1} \left( \frac{4-x}{2(x+1)} \right) + C \)
  • \( \frac{1}{\sqrt{5}} \cosh^{-1} \left( \frac{4+x}{2(x-1)} \right) + C \) 

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Step 1: Using a Suitable Substitution Let: \[ x = 2 \sinh t \] Differentiating both sides: \[ dx = 2\cosh t \, dt \] Rewriting the denominator: \[ \sqrt{x^2 + 4} = \sqrt{4\sinh^2 t + 4} = \sqrt{4(\sinh^2 t + 1)} = \sqrt{4\cosh^2 t} = 2\cosh t \] Thus, the integral transforms into: \[ \int \frac{dx}{(x+1)\sqrt{x^2+4}} = \int \frac{2\cosh t \, dt}{(2\sinh t + 1)2\cosh t} \]

Step 2: Evaluating the Integral The fraction simplifies to: \[ \int \frac{dt}{2\sinh t + 1} \] Using the inverse hyperbolic sine transformation: \[ I = \frac{1}{\sqrt{5}} \sinh^{-1} \left( \frac{4-x}{2(x+1)} \right) + C \] 

Was this answer helpful?
0
0