Question

Match the column

Answer 1

A) ${\int }_{-1}^1\frac{dx}{1+x^2}={\left[{\tan }^{-1}x\right]}_{-1}^1=(\frac{\pi }{4}+\frac{\pi }{4})=\frac{\pi }{2}$
B) ${\int }_0^1\frac{dx}{\sqrt{1-x^2}}={\left[{\sin }^{-1}x\right]}_0^1=(\frac{\pi }{2}-0)=\frac{\pi }{2}$
C) ${\int }_2^3\frac{dx}{1-x^2}={\left[\frac{1}{2}\log \frac{1+x}{1-x}\right]}_2^3=\frac{1}{2}(\log \frac{1+3}{1-3}-\log \frac{1+2}{1-2})=\frac{1}{2}\log \frac{4}{9}=\log \frac{2}{3}$
D) ${\int }_1^2\frac{dx}{\left|x\right|\sqrt{x^2-1}}={\left[{\sec }^{-1}x\right]}_1^2=({\sec }^{-1}2-{\sec }^{-1}1)=(\frac{\pi }{6}-\frac{\pi }{2})=-\frac{\pi }{3}$