QuestionProblem 2. (1 point)
Evaluate the integral $$\int_{1}^{\sqrt{6}} \frac{1}{1+x^{2}} d x$$

Studdy Solution
Recall that $\tan^{-1}(1) = \frac{\pi}{4}$ because $\tan(\frac{\pi}{4}) = 1$. The value of $\tan^{-1}(\sqrt{6})$ is an angle whose tangent is $\sqrt{6}$, which is not a standard angle, so it remains in its inverse tangent form:
$$\int_{1}^{\sqrt{6}} \frac{1}{1+x^2} \, dx = \tan^{-1}(\sqrt{6}) - \frac{\pi}{4}$$
The value of the definite integral is:
$$\tan^{-1}(\sqrt{6}) - \frac{\pi}{4}$$