(x-1)(x-2) can be written as x2-3x+2.
Therefore,
x2 - 3x + 2
= x2 - 3x +\(\frac 94\) -\(\frac 94\) + 2
=(x - \(\frac 32\))2 - \(\frac 14\)
=(x - \(\frac 32\))2 - (\(\frac 12\))2
∴ \(∫\)\(\frac {1}{\sqrt {(x-1)(x-2)}}\ dx\)= \(∫\frac {1}{\sqrt {(x-\frac 32)^2-(\frac 12)^2}} dx\)
Let x-\(\frac 32\) = t
∴ dx = dt
⇒ \(∫\frac {1}{\sqrt {(x-\frac 32)^2-(\frac 12)^2}} dx\) = \(∫\frac {1}{\sqrt {t^2-(\frac 12)^2}} dt\)
= \(log\ |t+\sqrt {t^2-(\frac 12)^2}|+C\)
=\(log\ |(x-\frac 32)+\sqrt {x^2-3x+2}|+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.
