Sketch first:

x
dy
x = y2 + 1

The curve x = y2 + 1, showing the portion "under" the curve from y = 1 to y = 5.

In this case, we express x as a function of y:

y=sqrt{x-1}

y^2=x-1

x=y^2+1

So the area is given by:

A=int_1^5(y^2+1) dy=[y^3/3+y]_1^5

=45 1/3\ text[sq units]

Note: For this particular example, we could have also summed it horizontally (integrating y and using dx), but we would need to break it up into sections first.