Question

${\int }_1^2\frac{dx}{x^2-2x+4}=$

• A. 0
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6\sqrt{3}}$

Answer 1

The correct option is D $\frac{\pi }{6\sqrt{3}}$

${\int }_1^2\frac{dx}{(x-1{)}^2+(\sqrt{3}{)}^2}=[\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{x-1}{\sqrt{3}}\right)+c]{\int }_1^2$
$=\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$=\frac{1}{\sqrt{3}}\cdot \frac{\pi }{6}$
$=\frac{\pi }{6\sqrt{3}}$