Problem
Simplify: \( \tan^{-1}(x + \sqrt{1 + x^2}), \quad x \in \mathbb{R} \)
Solution (Long Method)
Let:
\[ \theta = \tan^{-1}(x + \sqrt{1 + x^2}) \]
Then,
\[ \tan \theta = x + \sqrt{1 + x^2} \]
Taking reciprocal,
\[ \cot \theta = \frac{1}{x + \sqrt{1 + x^2}} \]
Rationalizing the denominator:
\[ \cot \theta = \frac{\sqrt{1 + x^2} – x}{(x + \sqrt{1 + x^2})(\sqrt{1 + x^2} – x)} \]
\[ = \frac{\sqrt{1 + x^2} – x}{1} = \sqrt{1 + x^2} – x \]
Let \( x = \tan \alpha \), then
\[ \sqrt{1 + x^2} = \sec \alpha \]
So,
\[ \tan \theta = \tan \alpha + \sec \alpha \]
Using identity:
\[ \tan\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) = \tan \alpha + \sec \alpha \]
Thus,
\[ \theta = \frac{\pi}{4} + \frac{\alpha}{2} \]
But \( \alpha = \tan^{-1} x \)
Final Answer
\[ \boxed{\frac{\pi}{4} + \frac{1}{2}\tan^{-1} x} \]