Question

If \[\int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} dx} = A\sqrt {{x^2} - 12x + 35} + \log \left| {x - 6 + \sqrt {{x^2} - 12x + 35} } \right| + C\], then what is the value of A.
(a). – 1
(b). \[\dfrac{1}{2}\]
(c). \[ - \dfrac{1}{2}\]
(d). 1

Answer 1

Hint:Multiply both numerator and denominator by \[\sqrt {x - 5} \] to simplify the integral. Use the formula \[\int {\dfrac{1}{{\sqrt {{{(x + b)}^2} - {a^2}} }}dx = \log |x + b + \sqrt {{{(x + b)}^2} - {a^2}} |} + C\] to solve the integral and then find the value of A.

Complete step-by-step answer:
Let us assign the given integral to the variable I, then we have:
  \[I = \int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} dx} \]
Multiply the numerator and denominator of the integrand by \[\sqrt {x - 5} \] to simplify the integral.
\[I = \int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} \times \sqrt {\dfrac{{x - 5}}{{x - 5}}} dx} \]
\[I = \int {\dfrac{{x - 5}}{{\sqrt {(x - 7)(x - 5)} }}dx} \]
Simplifying the denominator, we have:
\[I = \int {\dfrac{{x - 5}}{{\sqrt {{x^2} - 12x + 35} }}dx} ..........(1)\]
We can integrate the above integral using partial fractions. In partial fractions, we need to express the numerator in terms of the denominator and its derivatives. We can equate the numerator as a sum of the term in the square root of the denominator and its first derivative. We don’t consider the second derivative because the difference between the degree of numerator and denominator is one, hence, the terms until first derivative is considered.
\[x - 5 = a\dfrac{d}{{dx}}({x^2} - 12x + 35) + b\]
Simplifying, we have:
\[x - 5 = a(2x - 12) + b\]
\[x - 5 = 2ax - 12a + b\]
Comparing both sides of the equation, we have:
\[2a = 1\]
\[ - 12a + b = - 5\]
Solving for a, we have:
\[a = \dfrac{1}{2}...........(2)\]
Solving for b, we have:
\[ - 12\left( {\dfrac{1}{2}} \right) + b = - 5\]
\[ - 6 + b = - 5\]
\[b = - 5 + 6\]
\[b = 1..........(3)\]
Using equation (2) and equation (3), we can write equation (1) as follows:
\[I = \dfrac{1}{2}\int {\dfrac{{2x - 12}}{{\sqrt {{x^2} - 12x + 35} }}dx} + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}dx} \]
We can express the integral as the sum of two integrals as follows:
\[I = {I_1} + {I_2}..........(4)\]
Let’s solve the first integral.
\[{I_1} = \dfrac{1}{2}\int {\dfrac{{2x - 12}}{{\sqrt {{x^2} - 12x + 35} }}dx} \]
We can solve this integral by substitution of variables method.
Let \[t = {x^2} - 12x + 35\].
Differentiating both sides, we have:
\[dt = 2x - 12x\]
Then, we have the integral as follows:
\[{I_1} = \dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt t }}} \]
We know that the integral of \[\dfrac{1}{{\sqrt x }}\] is \[2\sqrt x \], then, we have:
\[{I_1} = \dfrac{1}{2}.2\sqrt t \]
\[{I_1} = \sqrt t + {C_1}\]
Substituting the value of t, we have:
\[{I_1} = \sqrt {{x^2} - 12x + 35} + {C_1}...........(5)\]
Now, let us solve the second integral.
\[{I_2} = \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}dx} \]
Let us use completing the square method to write the denominator in terms of squares.
\[{I_2} = \int {\dfrac{1}{{\sqrt {{x^2} - 2.6.x + 36 - 1} }}dx} \]
Simplifying, we have:
\[{I_2} = \int {\dfrac{1}{{\sqrt {{{(x - 6)}^2} - {1^2}} }}dx} \]
We can use the following formula to solve the integral.
\[\int {\dfrac{1}{{\sqrt {{{(x + b)}^2} - {a^2}} }}dx = \log |x + b + \sqrt {{{(x + b)}^2} - {a^2}} |} + C\]
Hence, we have the following:
\[{I_2} = \log |x - 6 + \sqrt {{{(x - 6)}^2} - {1^2}} | + {C_2}\]
Simplifying, we have:
\[{I_2} = \log |x - 6 + \sqrt {{x^2} - 12x + 36 - 1} | + {C_2}\]
\[{I_2} = \log |x - 6 + \sqrt {{x^2} - 12x + 35} | + {C_2}........(6)\]
Using equation (5) and equation (6) in equation (4), we have:
\[I = \sqrt {{x^2} - 12x + 35} + \log |x - 6 + \sqrt {{x^2} - 12x + 35} | + {C_1} + {C_2}\]
Comparing the result with the question, the value of A is 1.
Hence, the correct answer is option (d).

Note: You might miss the \[\dfrac{1}{2}\] factor in the first integral. You may also write the integral of \[\dfrac{1}{{\sqrt x }}\] is \[\dfrac{{\sqrt x }}{2}\], it is wrong. The integral of \[\dfrac{1}{{\sqrt x }}\] is \[2\sqrt x \].Students should remember the important integration formulas for solving these types of problems.