To integrate the expression 1/√(x-a)(x-b) with respect to x, we can use a method called partial fraction decomposition. This technique allows us to express the given expression as a sum of simpler fractions that we can integrate individually.



To begin, we factor the denominator (x-a)(x-b) to get two linear factors. The partial fraction decomposition of 1/√(x-a)(x-b) is given by:

1/√(x-a)(x-b) = A/√(x-a) + B/√(x-b)

Here, A and B are constants that we need to find.



To find the values of A and B, we need to equate the numerators of the fractions in the partial fraction decomposition to the original numerator, which is 1. This gives us the following equation:

1 = A√(x-b) + B√(x-a)

To simplify the equation, we can square both sides:

1 = A^2(x-b) + 2AB√(x-a)(x-b) + B^2(x-a)

Since the equation holds for all values of x, the coefficients of each power of x must be equal. This leads to the following system of equations:

A^2 + B^2 = 0 (coefficient of x^0)
2AB = 0 (coefficient of x^1)

From the second equation, we can see that either A or B must be zero. However, since A^2 + B^2 = 0, both A and B must be zero.

Therefore, we conclude that A = 0 and B = 0.



With A = 0 and B = 0, the partial fraction decomposition becomes:

1/√(x-a)(x-b) = 0/√(x-a) + 0/√(x-b) = 0

The integral of 0 with respect to x is simply a constant, so the final result of the integration is:

∫ 0 dx = C

Here, C represents the constant of integration.



Therefore, the integral of 1/√(x-a)(x-b) with respect to x is C, where C is the constant of integration.