Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 63

We first need to find the equation of the tangent line. We will do that using the following formula:
y=y_1+f'(x_1)(x-x_1)
where (x_1,y_1) is the point where the tangent touches the graph.
First we calculate the derivative.
f'(x)=(0cdot(x^2+1)-1cdot2x)/(x^2+1)^2=-(2x)/(x^2+1)^2
f(1)=-2/2^2=-1/2
Now we can calculate the equation of the tangent line.
y=1/2-1/2(x-1)
y=1/2-1/2x+1/2
y=1-1/2x
We have to determine the bounds of integration. To do that, we need to know the where the tangent line intersects the graph of the function (point A in the image below). Hence, we need to solve the following system of equations.
y=1/(x^2+1)
y=1-1/2x
1-1/2x=1/(x^2+1)
Multiply by 2(x^2+1).
2x^2+2-x^3-x=2
-x^3+2x^2-x=0
Factoring yields.
-x(x^2-2x+1)=0
From this we see that the solutions are x_1=0, x_(2,3)=1.
Therefore, the bounds of integration will be 0 an 1. To set up the integral we only need to look at the image below and see that the tangent line lies below the graph of the function which means we need to subtract it from the function. Alternatively, we could switch the bounds of integration or simply take absolute value from the whole integral.
A=int_0^1 (1/(x^2+1)-1+1/2x)dx=
(arctan x-x+x^2/4)|_0^1=
pi/4-1+1/4-0+0-0=1/4(pi-3)
The are of the region bounded by the graph of the given function and tangent to that function at (1,1/2) is equal to 1/4(pi-3).