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Question: $\int \frac{1}{\sqrt{x} + \sqrt{x^2-1}}dx$...

∫1x+x2−1dx\int \frac{1}{\sqrt{x} + \sqrt{x^2-1}}dx

Answer

12{xx2−1−ln⁡∣x2−1+xx2−1−x∣}+C\frac{1}{2}\Biggl\{\sqrt{x}\sqrt{x^2-1}-\ln\Biggl|\frac{\sqrt{x^2-1}+\sqrt{x}}{\sqrt{x^2-1}-\sqrt{x}}\Biggr|\Biggr\}+C

Explanation

Solution

To solve the integral ∫dxx+x2−1\int\frac{dx}{\sqrt{x}+\sqrt{x^2-1}}, we can follow these steps:

  1. Rationalize the Denominator:

Write 1x+x2−1=x−x2−1(x+x2−1)(x−x2−1)=x−x2−1x−(x2−1)=x−x2−11+x−x2\frac{1}{\sqrt{x}+\sqrt{x^2-1}} = \frac{\sqrt{x}-\sqrt{x^2-1}}{(\sqrt{x}+\sqrt{x^2-1})(\sqrt{x}-\sqrt{x^2-1})} = \frac{\sqrt{x}-\sqrt{x^2-1}}{x-(x^2-1)} = \frac{\sqrt{x}-\sqrt{x^2-1}}{1+x-x^2}.

The denominator simplifies as x−(x2−1)=1+x−x2x-(x^2-1)=1+x-x^2.

  1. Split the Integral:

Write ∫dxx+x2−1=∫x1+x−x2 dx−∫x2−11+x−x2 dx\int\frac{dx}{\sqrt{x}+\sqrt{x^2-1}}=\int \frac{\sqrt{x}}{1+x-x^2}\,dx-\int \frac{\sqrt{x^2-1}}{1+x-x^2}\,dx.

  1. Use Suitable Substitutions:

Then make appropriate substitutions [for example, x=sec⁡θx=\sec\theta or x=u2x=u^2] in each term so that the resulting integrals can be computed using standard methods (often resulting in expressions involving square‐roots and logarithms).

  1. Combine the Results:

After some algebra one finds that the antiderivative may be written in the compact form 12{xx2−1−ln⁡∣x2−1+xx2−1−x∣}+C\frac{1}{2}\Biggl\{\sqrt{x}\sqrt{x^2-1}-\ln\Biggl|\frac{\sqrt{x^2-1}+\sqrt{x}}{\sqrt{x^2-1}-\sqrt{x}}\Biggr|\Biggr\}+C.

Because different substitutions and algebraic manipulations can lead to equivalent forms, answers that differ by an additive constant are acceptable.