Question

Find: \[ I = \int \frac{\sqrt{x}}{(x+1)(x-1)} \, dx \]

Hint:

For integrals involving square roots, use substitution to simplify the expression and apply partial fraction decomposition when necessary.

Answer 1

Step 1: Substitute \( \sqrt{x} = t \)
Let \( \sqrt{x} = t \), so \( x = t^2 \) and \( dx = 2t \, dt \). Substituting into the integral: \[ I = \int \frac{t}{(t^2 + 1)(t^2 - 1)} \cdot 2t \, dt = \int \frac{2t^2}{(t^2 + 1)(t^2 - 1)} \, dt. \] Step 2: Simplify the integrand
Rewrite \( \frac{2t^2}{(t^2 + 1)(t^2 - 1)} \) using partial fraction decomposition. We have: \[ \frac{2t^2}{(t^2 + 1)(t^2 - 1)} = \frac{1}{t^2 + 1} + \frac{1}{t^2 - 1}. \] Step 3: Integrate each term
The integral becomes: \[ I = \int \frac{1}{t^2 + 1} \, dt + \int \frac{1}{t^2 - 1} \, dt. \] 1. For \( \int \frac{1}{t^2 + 1} \, dt \), the result is: \[ \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t). \] 2. For \( \int \frac{1}{t^2 - 1} \, dt \), rewrite using partial fractions: \[ \frac{1}{t^2 - 1} = \frac{1}{2} \left( \frac{1}{t - 1} - \frac{1}{t + 1} \right). \] Integrating term by term: \[ \int \frac{1}{t^2 - 1} \, dt = \frac{1}{2} \ln|t - 1| - \frac{1}{2} \ln|t + 1| = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right|. \] Step 4: Combine results and back-substitute
Substitute back \( t = \sqrt{x} \): \[ I = \tan^{-1}(\sqrt{x}) + \frac{1}{2} \ln \left| \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right| + C. \] 
Conclusion: The solution to the integral is: \[ I = \tan^{-1}(\sqrt{x}) + \frac{1}{2} \ln \left| \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right| + C. \]