Question

${\int }_{-1}^3(Ta{n}^{-1}\frac{x}{x^2+1}+Ta{n}^{-1}\frac{x^2+1}{x})dx=$

• A. $\pi$
• B. $2\pi$
• C. $4\pi$
• D. $none$

Answer 1

The correct option is A $\pi$

${\int }_{-1}^3(Ta{n}^{-1}\frac{x}{x^2+1}+Ta{n}^{-1}\frac{x^2+1}{x})dx={\int }_{-1}^0[Ta{n}^{-1}\frac{x}{x^2+1}+Ta{n}^{-1}\frac{x^2+1}{x}]dx+{\int }_0^3[Ta{n}^{-1}\frac{x}{x^2+1}+Ta{n}^{-1}\frac{x^2+1}{x}]dx={\int }_{-1}^0-\frac{\pi }{2}dx+{\int }_0^3\frac{\pi }{2}dx=-\frac{\pi }{2}+\frac{\pi }{2}(3-(0\left)\right)=\pi$