Question

Solve $\int {\dfrac{1}{{x(x + 1)}}dx = }$
A. $\ln |\dfrac{{x + 1}}{x}| + c$
B. $\ln |\dfrac{x}{{x + 1}}| + c$
C. $\ln |\dfrac{{x - 1}}{x}| + c$
D. $\ln |\dfrac{{x - 1}}{{x + 1}}| + c$

Answer 1

Now, this question is based on integration by partial fractions.According to the integration by partial fractions, we want to evaluate a proper rational fraction. In such cases, it is possible to write the integrand as a sum of simpler rational functions using partial fraction decomposition.

Formula used:
The form of the partial fraction used in this question is shown below,
$\dfrac{1}{{(x - a)(x - b)}} = \dfrac{A}{{(x - a)}} + \dfrac{B}{{(x - b)}}$
where $a$ and $b$ are constants.
The integration used in solving this question is shown below,
$\int {\dfrac{{dx}}{x} = \ln |x|} + c$
where $c$ is an arbitrary constant.

Complete step by step answer:
Let us start the question by using the integration by using partial fractions
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{A}{x} + \dfrac{B}{{x + 1}}$
Here $A$ and $B$ are constants.
Now, solving the denominator on the right-hand side of the equation as shown below
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{{A(x + 1) + Bx}}{{x(x + 1)}}$
Now, the denominators on both the sides of the equation are equal; therefore, their numerators must also be equal,
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{{x(A + B) + A}}{{x(x + 1)}}$

Now, there is no parameter x in the numerator on the left-hand side of the equation.Therefore, the coefficient of x in the numerator on the right-hand side of the equation must also be equal to zero,
$\Rightarrow A + B = 0$
$\Rightarrow A = - B - - - - (i)$
Now, the constant in the numerator on the left-hand side of the equation is one, and thus, the value of the constant in the numerator on the right-hand side of the equation is also equal to one.
$\Rightarrow A = 1$
From the equation, $(i)$ we can calculate the value of B as shown below,
$\Rightarrow B = - 1$

Now, putting these values in the equation shown below
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{A}{x} + \dfrac{B}{{x + 1}}$
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{1}{x} + \dfrac{{( - 1)}}{{x + 1}}$
$\Rightarrow \dfrac{1}{{x(x + 1)}} = \dfrac{1}{x} - \dfrac{1}{{x + 1}}$
Now, integrating this function as shown below,
$\Rightarrow \int {\dfrac{1}{{x(x + 1)}}dx = \int {(\dfrac{1}{x} - \dfrac{1}{{x + 1}})} dx}$
$\Rightarrow \int {(\dfrac{1}{x} - \dfrac{1}{{x + 1}})} dx = \int {\dfrac{1}{x}dx - \int {\dfrac{1}{{x + 1}}} } dx$
By using the basic integration properties, we can integrate the above function
$\Rightarrow \int {\dfrac{1}{x}dx - \int {\dfrac{1}{{x + 1}}} } dx = \ln |x| - \ln |x + 1| + c$, where $c$ is an arbitrary constant
$\therefore \int {\dfrac{1}{x}dx - \int {\dfrac{1}{{x + 1}}} } dx = \ln |\dfrac{x}{{x + 1}}| + c$, where $c$ is an arbitrary constant

Thus, option B is the correct answer.

Note: We can see many complex rational expressions. If we try to solve the problems in the complex form, it will take a lot of time to find the solution. To avoid this complexity, we have to continue the problem by reducing the complex form of the rational expression into the simpler form. Partial fraction decomposition is one of the methods, which is used to decompose the rational expressions into simpler partial fractions. This process is more useful in the integration process.