Question

The value of ${\int }_0^{\infty }\frac{1}{1+x^2}dx+{\int }_0^{\infty }\frac{sinx}{x}dx$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{3\pi }{2}$
• D. $1$

Answer 1

The correct option is B $\pi$

${\int }_0^{\infty }\frac{1}{1+x^2}dx+{\int }_0^{\infty }\frac{sinx}{x}dx={\left[ta{n}^{-1}x\right]}_0^{\infty }+{\int }_0^{\infty }\frac{sinx}{x}dx$
$[$ Using Laplace transform method, we know that
${\int }_0^{\infty }\frac{sinx}{x}dx=\frac{\pi }{2}=\frac{\pi }{2}+\frac{\pi }{2}=\pi$