Question

If $\int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} dx = A\sqrt {{x^2} - 12x + 35} + \log |x - 6 + \sqrt {{x^2} - 12x + 35} | + C}$ then A is equal to,
A. $- 1$
B. $\dfrac{1}{2}$
C. $\dfrac{{ - 1}}{2}$
D. $1$

Answer 1

Take the given expression in the integral and find the partial fractions. Now write the numerator in the form of differentiation of denominator and then integrate. Do the same with the other term. Finally, compare the equation we got with the expression in the question to get the value of $A$.

Formula used: $\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = \log |x + \sqrt {{x^2} - {a^2}} |} + C$

Complete step-by-step solution:
The given integral that needs to be solved is $\int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} } dx$
Firstly, we multiply the numerator and denominator with $(x - 5)$
$\Rightarrow \int {\sqrt {\dfrac{{x - 5}}{{x - 7}} \times \dfrac{{x - 5}}{{x - 5}}} } dx$
Now club the two same expressions in the numerator and the square of it.
$\Rightarrow \int {\sqrt {\dfrac{{{{(x - 5)}^2}}}{{(x - 7)(x - 5)}}} } dx$
Multiply the two linear expressions in the denominator
$\Rightarrow \int {\sqrt {\dfrac{{{{(x - 5)}^2}}}{{{x^2} - 12x + 35}}} } dx$
The square in the numerator gets canceled with the square root.
The square root in the denominator remains the same.
$\Rightarrow \int {\dfrac{{x - 5}}{{\sqrt {{x^2} - 12x + 35} }}} dx$
Divide and multiply the expression with the constant $2$
$\Rightarrow \dfrac{1}{2}\int {\dfrac{{2(x - 5)}}{{\sqrt {{x^2} - 12x + 35} }}} dx$
$\Rightarrow \dfrac{1}{2}\int {\dfrac{{2x - 10}}{{\sqrt {{x^2} - 12x + 35} }}} dx$
Now split the expression in such a way that the differentiation of the denominator is the numerator.
$\Rightarrow \dfrac{1}{2}\int {\dfrac{{2x - 12}}{{\sqrt {{x^2} - 12x + 35} }}dx - \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} } dx$
Let’s now consider ${x^2} - 12x + 35 = t$
On differentiating both sides,
$\Rightarrow (2x - 12)dx = dt$
$\Rightarrow 2x - 12 = \dfrac{{dt}}{{dx}}$
Now substitute these in the first term.
$\Rightarrow \dfrac{1}{2}\int {\dfrac{{\dfrac{{dt}}{{dx}}}}{{\sqrt t }}dx - \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} } dx$
After simplifying we get,
$\Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{\sqrt t }}dt - \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} } dx$
The square root can be written in terms of power as,
$\Rightarrow \dfrac{1}{2}\int {{{(t)}^{\dfrac{{ - 1}}{2}}}dt - \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} } dx$
Now upon integrating the first term we get,
$\Rightarrow \dfrac{1}{2}\left[ {\dfrac{{{t^{\dfrac{{ - 1}}{2} + 1}}}}{{\dfrac{{ - 1}}{2} + 1}}} \right] + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} dx$
$\Rightarrow \dfrac{1}{2}\left[ {\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right] + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} dx$
Now write the power of $t$ back into square root form.
$\Rightarrow \dfrac{1}{2}\left[ {\dfrac{{\sqrt t }}{{\dfrac{1}{2}}}} \right] + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} dx$
On further simplification in the first term, we get,
$\Rightarrow \sqrt t + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35} }}} dx$
Now add $1$ and subtract $1$ in the denominator of the second term.
$\Rightarrow \sqrt t + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 35 + 1 - 1} }}} dx$
Evaluate to get,
$\Rightarrow \sqrt t + \int {\dfrac{1}{{\sqrt {{x^2} - 12x + 36 - 1} }}} dx$
The denominator can also be written as,
$\Rightarrow \sqrt t + \int {\dfrac{1}{{\sqrt {{{(x - 6)}^2} - 1} }}} dx$
Apply square on $1$ also.
$\Rightarrow \sqrt t + \int {\dfrac{1}{{\sqrt {{{(x - 6)}^2} - {1^2}} }}} dx$
Now, this is the form of the formula, $\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = \log |x + \sqrt {{x^2} - {a^2}} |} + C$
Put the values in the formula to get,
$\Rightarrow \sqrt t + \log |x - 6 + \sqrt {{x^2} - 12x + 35} |$
Now substitute the value of assumed $t$ back in the equation.
$\Rightarrow \sqrt {{x^2} - 12x + 35} + \log |x - 6 + \sqrt {{x^2} - 12x + 35} | + C$
Now compare our equation with the one given in the question.
$\int {\sqrt {\dfrac{{x - 5}}{{x - 7}}} dx = A\sqrt {{x^2} - 12x + 35} + \log |x - 6 + \sqrt {{x^2} - 12x + 35} | + C}$
We can now see that the first term is the same as that of the given expression in the question so $A = 1$

Option D is the correct answer.

Note: Memorize the integral of quadratic denominators formula to easily get the equations into the same form and then substituting in the formula. Also, make the numerator as the differentiation form of the denominator to easily evaluate.