The area of the region bounded by the curve $C:y=\frac{x+1}{x^2+1}$ and the line $y = 1$, is
Answer & explanation
Correct answer: option 3
Let A be the required area. Then,
$A=\int\limits_0^1(y-1)dx=\int\limits_0^1\left(\frac{x+1}{x^2+1}-1\right)dx$
$⇒A=\int\limits_0^1\left[\frac{1}{2}\log(x^2+1)+\tan^2x-x\right]_0^1=\frac{1}{2}ln\,2+\frac{π}{4}-1$