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

∫dxx2+1+x2−1\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2-1}}

Answer

x4(x2+1−x2−1)+14(ln⁡∣x+x2+1∣+ln⁡∣x+x2−1∣)+C\frac{x}{4}(\sqrt{x^2+1} - \sqrt{x^2-1}) + \frac{1}{4}(\ln|x+\sqrt{x^2+1}| + \ln|x+\sqrt{x^2-1}|) + C

Explanation

Solution

To evaluate the integral ∫dxx2+1+x2−1\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2-1}}, we first rationalize the denominator.

Step 1: Rationalize the Denominator

Multiply the numerator and the denominator by the conjugate of the denominator, which is x2+1−x2−1\sqrt{x^2+1}-\sqrt{x^2-1}.

1x2+1+x2−1=1x2+1+x2−1×x2+1−x2−1x2+1−x2−1\frac{1}{\sqrt{x^2+1}+\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2+1}+\sqrt{x^2-1}} \times \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}

Using the identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2 in the denominator:

=x2+1−x2−1(x2+1)2−(x2−1)2= \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{(\sqrt{x^2+1})^2 - (\sqrt{x^2-1})^2} =x2+1−x2−1(x2+1)−(x2−1)= \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{(x^2+1) - (x^2-1)} =x2+1−x2−1x2+1−x2+1= \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{x^2+1-x^2+1} =x2+1−x2−12= \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{2}

Step 2: Integrate the Simplified Expression

Now, substitute this simplified expression back into the integral:

∫dxx2+1+x2−1=∫x2+1−x2−12dx\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2-1}} = \int \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{2} dx =12[∫x2+1dx−∫x2−1dx]= \frac{1}{2} \left[ \int \sqrt{x^2+1} dx - \int \sqrt{x^2-1} dx \right]

Step 3: Apply Standard Integration Formulas

We use the standard integration formulas:

  1. ∫x2+a2dx=x2x2+a2+a22ln⁡∣x+x2+a2∣+C\int \sqrt{x^2+a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}| + C
  2. ∫x2−a2dx=x2x2−a2−a22ln⁡∣x+x2−a2∣+C\int \sqrt{x^2-a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}| + C

For the first integral, ∫x2+1dx\int \sqrt{x^2+1} dx, we have a=1a=1:

∫x2+1dx=x2x2+1+122ln⁡∣x+x2+1∣=x2x2+1+12ln⁡∣x+x2+1∣\int \sqrt{x^2+1} dx = \frac{x}{2}\sqrt{x^2+1} + \frac{1^2}{2}\ln|x+\sqrt{x^2+1}| = \frac{x}{2}\sqrt{x^2+1} + \frac{1}{2}\ln|x+\sqrt{x^2+1}|

For the second integral, ∫x2−1dx\int \sqrt{x^2-1} dx, we have a=1a=1:

∫x2−1dx=x2x2−1−122ln⁡∣x+x2−1∣=x2x2−1−12ln⁡∣x+x2−1∣\int \sqrt{x^2-1} dx = \frac{x}{2}\sqrt{x^2-1} - \frac{1^2}{2}\ln|x+\sqrt{x^2-1}| = \frac{x}{2}\sqrt{x^2-1} - \frac{1}{2}\ln|x+\sqrt{x^2-1}|

Step 4: Combine the Results

Substitute these back into the expression from Step 2:

12[(x2x2+1+12ln⁡∣x+x2+1∣)−(x2x2−1−12ln⁡∣x+x2−1∣)]+C\frac{1}{2} \left[ \left( \frac{x}{2}\sqrt{x^2+1} + \frac{1}{2}\ln|x+\sqrt{x^2+1}| \right) - \left( \frac{x}{2}\sqrt{x^2-1} - \frac{1}{2}\ln|x+\sqrt{x^2-1}| \right) \right] + C =12[x2x2+1+12ln⁡∣x+x2+1∣−x2x2−1+12ln⁡∣x+x2−1∣]+C= \frac{1}{2} \left[ \frac{x}{2}\sqrt{x^2+1} + \frac{1}{2}\ln|x+\sqrt{x^2+1}| - \frac{x}{2}\sqrt{x^2-1} + \frac{1}{2}\ln|x+\sqrt{x^2-1}| \right] + C

Distribute the 12\frac{1}{2}:

=x4x2+1−x4x2−1+14ln⁡∣x+x2+1∣+14ln⁡∣x+x2−1∣+C= \frac{x}{4}\sqrt{x^2+1} - \frac{x}{4}\sqrt{x^2-1} + \frac{1}{4}\ln|x+\sqrt{x^2+1}| + \frac{1}{4}\ln|x+\sqrt{x^2-1}| + C

This can also be written by factoring out common terms:

=x4(x2+1−x2−1)+14(ln⁡∣x+x2+1∣+ln⁡∣x+x2−1∣)+C= \frac{x}{4}(\sqrt{x^2+1} - \sqrt{x^2-1}) + \frac{1}{4}(\ln|x+\sqrt{x^2+1}| + \ln|x+\sqrt{x^2-1}|) + C

The final answer is x4(x2+1−x2−1)+14(ln⁡∣x+x2+1∣+ln⁡∣x+x2−1∣)+C\frac{x}{4}(\sqrt{x^2+1} - \sqrt{x^2-1}) + \frac{1}{4}(\ln|x+\sqrt{x^2+1}| + \ln|x+\sqrt{x^2-1}|) + C.

Explanation of the solution: The integral is simplified by rationalizing the denominator, which transforms the integrand into a difference of two standard forms (x2+a2\sqrt{x^2+a^2} and x2−a2\sqrt{x^2-a^2}). Each part is then integrated using their respective standard formulas, and the results are combined.