(x-a)(x-b) can be written as x2 - (a+b)x + ab.
Therefore,
x2- (a+b)x + ab
= x2- (a+b)x + \(\frac {(a+b)^2}{4}\) - \(\frac {(a+b)^2}{4}\) + ab
= [x-(\(\frac {a+b}{2}\))]2 - \(\frac {(a-b)^2}{4}\)
⇒ \(∫\)\(\frac {1}{\sqrt {(x-a)(x-b)}}\ dx\) = \(∫\frac {1}{\sqrt {{x-(\frac {a+b}{2})}^2-(\frac {a-b}{2})^2}} dx\)
Let x - (\(\frac {a+b}{2}\)) = t
∴ dx = dt
⇒ \(∫\frac {1}{\sqrt {{x-(\frac {a+b}{2})}^2-(\frac {a-b}{2})^2}} dx\) = \(∫\frac {1}{\sqrt {t^2-(\frac {a-b}{2})^2}}dt\)
= \(log \ |t+\sqrt {t^2-(\frac {a-b}{2})^2|}+C\)
= \(log \ |{x-(\frac {a+b}{2})}+\sqrt {(x-a)(x-b)}|+C\)
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 ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?
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.
