Question

Evaluate the integral $\smallint \dfrac{{dx}}{{(x - 1)\sqrt {{x^2} - 1} }} =$
A). $- \sqrt {\dfrac{{x - 1}}{{x + 1}}} + C$
B). $\sqrt {\dfrac{{x - 1}}{{{x^2} + 1}}} + C$
C). $- \sqrt {\dfrac{{x + 1}}{{x - 1}}} + C$
D). $\sqrt {\dfrac{{{x^2} + 1}}{{x - 1}}} + C$

Answer 1

The above integral is integrated by using a substitution method. The given integral is of the form $\smallint \dfrac{{dx}}{{p\sqrt q }}$ where in this question $p$ is linear and $q$ is quadratic. Using the substitution method, we will substitute $p = \dfrac{1}{t}$.

Complete step-by-step solution:
The given integral is $\smallint \dfrac{{dx}}{{(x - 1)\sqrt {{x^2} - 1} }}$
Here $p = (x - 1)$ and $q = {x^2} - 1$
Using substitution method,
We substitute $p = \dfrac{1}{y}$ ,
i.e., $x - 1 = \dfrac{1}{y}$
Taking $- 1$ to the other side,
$x = 1 + \dfrac{1}{y}$
Differentiating this equation with respect to $x$,
$dx = - \dfrac{1}{{{y^2}}}dy$
Substituting the values of $x$ and $dx$ in the given integral,
$= \smallint \dfrac{1}{{\dfrac{1}{y}\sqrt {{{\left( {\dfrac{1}{y} + 1} \right)}^2} - 1} }} \times \left( {\dfrac{{ - 1}}{{{y^2}}}} \right)dy$
Taking square of $1 + \dfrac{1}{y}$ ,
$= \smallint \dfrac{1}{{\dfrac{1}{y}\sqrt {\dfrac{1}{{{y^2}}} + 2 \times \dfrac{1}{y} + 1 - 1} }} \times \left( {\dfrac{{ - 1}}{{{y^2}}}} \right)dy$
Canceling $y$ from numerator and denominator,
$= \smallint \dfrac{1}{{\sqrt {\dfrac{1}{{{y^2}}} + 2 \times \dfrac{1}{y} + 1 - 1} }} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
Solving the terms inside the square root,
$= \smallint \dfrac{1}{{\sqrt {\dfrac{1}{{{y^2}}} + \dfrac{2}{y}} }} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
Multiplying and dividing $y$ to the term $\dfrac{2}{y}$ ,
$= \smallint \dfrac{1}{{\sqrt {\dfrac{1}{{{y^2}}} + \dfrac{{2y}}{{{y^2}}}} }} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
Since the denominator is same, we can add the numerator term inside the square root,
$= \smallint \dfrac{1}{{\sqrt {\dfrac{{1 + 2y}}{{{y^2}}}} }} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
We can write the equation as,
$= \smallint \dfrac{1}{{\dfrac{{\sqrt {1 + 2y} }}{{\sqrt {{y^2}} }}}} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
Taking square root of ${y^2}$,
$= \smallint \dfrac{1}{{\dfrac{{\sqrt {1 + 2y} }}{y}}} \times \left( {\dfrac{{ - 1}}{y}} \right)dy$
Canceling the $y$ from numerator and denominator,
$= \smallint \dfrac{{ - 1}}{{\sqrt {1 + 2y} }}dy$
Taking the square root term to the numerator,
$= - \smallint {(1 + 2y)^{\dfrac{{ - 1}}{2}}}dy$
This integral is of the form $\smallint {(ax + b)^n} = \dfrac{{{{(ax + b)}^{n + 1}}}}{{a(n + 1)}} + c$,
Integrating the above equation according to the formula,
$= \dfrac{{ - {{(1 + 2y)}^{\dfrac{{ - 1}}{2} + 1}}}}{{2\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + c$
Solving the denominator ,
$= \dfrac{{ - {{(1 + 2y)}^{\dfrac{{ - 1}}{2} + 1}}}}{{2\left( {\dfrac{1}{2}} \right)}} + c$
Canceling $2$ from numerator and denominator,
$= \dfrac{{ - {{(1 + 2y)}^{\dfrac{{ - 1}}{2} + 1}}}}{1} + c$
Solving the power of numerator term,
$= - {(1 + 2y)^{\dfrac{1}{2}}} + c$
We can write the equation as,
$= - \sqrt {1 + 2y} + c$
Substituting the value of $y = \dfrac{1}{{x - 1}}$,
$= - \sqrt {1 + 2\left( {\dfrac{1}{{x - 1}}} \right)} + c$
Multiplying and dividing $1$ by $x - 1$,
$= - \sqrt {\dfrac{{x - 1}}{{x - 1}} + 2\left( {\dfrac{1}{{x - 1}}} \right)} + c$
Since the denominator is same, we can add the numerator term,
$= - \sqrt {\dfrac{{x - 1 + 2}}{{x - 1}}} + c$
Solving the numerator part,
$= - \sqrt {\dfrac{{x + 1}}{{x - 1}}} + c$
Therefore, $\smallint \dfrac{{dx}}{{(x - 1)\sqrt {{x^2} - 1} }} = - \sqrt {\dfrac{{x + 1}}{{x - 1}}} + c$
The correct option is C. $- \sqrt {\dfrac{{x + 1}}{{x - 1}}} + C$.

Note: While canceling the terms from numerator and denominator, one should be careful while doing that as in integration many terms are in the denominator of the denominator and after doing integration, one should not forget to add the constant term as in above problem because no limits were given, constant is a must.