The area of the region bounded by the curve $C:y=\frac{x+1}{x^2+1}$ and the line $y = 1$, is

$m1-\frac{1}{2}ln\,2+\frac{π}{4}$ $ln\,2-\frac{π}{4}+1$ $\frac{1}{2}ln\,2+\frac{π}{4}-1$ $ln\,2-\frac{π}{2}+1$

$\frac{1}{2}ln\,2+\frac{π}{4}-1$

Let A be the required area. Then, $A=\int\limits_0^1(y-1)dx=\int\limits_0^1\left(\frac{x+1}{x^2+1}-1\right)dx$ $⇒A=\int\limits_0^1\left[\frac{1}{2}\log(x^2+1)+\tan^2x-x\right]_0^1=\frac{1}{2}ln\,2+\frac{π}{4}-1$