Question

$\underset{1}{\overset{\sqrt{3}}{\int }}\frac{1}{1+x^2}dx$ is equal to

(a) $\frac{\mathrm{\pi }}{12}$

(b) $\frac{\mathrm{\pi }}{6}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) $\frac{\mathrm{\pi }}{3}$

Answer 1

(a) $\frac{\mathrm{\pi }}{12}$

$$\begin{gathered} {\int }_1^{\sqrt{3}}\frac{1}{1+x^2}dx \\ ={\left[{\tan }^{-1}x\right]}_1^{\sqrt{3}} \\ =\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{12} \end{gathered}$$
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