Question

Evaluate the integral $\dfrac{{\log (x + 1) - \log x}}{{x(x + 1)}}dx$ .

Answer 1

Hint: We solve the required integral by the method of substitution. Here we have to find the required integration by first substituting the numerator as $t$ and then we have to find $dx$ . Upon substituting the values we will see that the integration reduces to a simple term. This will make the process of finding the integration smoother.

Formula used: We know that the derivative of $\log x$ is $\dfrac{1}{x}$ provided $x \ne 0$ .
The integration of ${x^n}$ is $\int {{x^n}dx} = \dfrac{{{x^{(n + 1)}}}}{{n + 1}} + c,n \ne - 1$ .

Complete step-by-step answer:
We are given the integral $\dfrac{{\log (x + 1) - \log x}}{{x(x + 1)}}dx$ .
We need to find the required indefinite integral.
So first we substitute the numerator as $t$ in the given integral.
Let $t = \log (x + 1) - \log x$ .
The required integration becomes $\int {\dfrac{t}{{x(x + 1)}}dx} $ .
Next, we differentiate the numerator.
Differentiating both sides of $t = \log (x + 1) - \log x$ w.r.t. $x$ we get:
$
\dfrac{{dt}}{{dx}} = \left( {\dfrac{1}{{(x + 1)}} - \dfrac{1}{x}} \right) \\
\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{x - (x + 1)}}{{x(x + 1)}} \\
\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{x - x - 1}}{{x(x + 1)}} \\
\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{ - 1}}{{x(x + 1)}} \\
\Rightarrow dt = \dfrac{{ - dx}}{{x(x + 1)}} \\
$
Now we substitute the value of $\dfrac{{dx}}{{x(x + 1)}} = - dt$ in the given integral.
Let $I = \int {\dfrac{{\left( {\log (x + 1) - \log x} \right)}}{{x(x + 1)}}} dx$
$
\therefore I = \int {\dfrac{{\left( {\log (x + 1) - \log x} \right)}}{{x(x + 1)}}} dx \\
\Rightarrow I = \int {\dfrac{t}{{x(x + 1)}}dx} \\
\Rightarrow I = \int {t \times ( - dt)} \\
\Rightarrow I = - \int {tdt} \\
\Rightarrow I = - \dfrac{{{t^2}}}{2} + c \\
$
Here, $c$ is known as the constant of integration.
Replacing the value of $t = \log (x + 1) - \log x$ in the equation we get:
$
I = - \dfrac{{{{(\log (x + 1) - \log x)}^2}}}{2} + c \\
\Rightarrow I = - \dfrac{{\log {{\left( {\dfrac{{x + 1}}{x}} \right)}^2}}}{2} + c \\
$
Here we used the property of $\log a - \log b = \log \dfrac{a}{b}$ .
Therefore the required value of the integral $\dfrac{{\log (x + 1) - \log x}}{{x(x + 1)}}dx$ is $ - \dfrac{{\log {{\left( {\dfrac{{x + 1}}{x}} \right)}^2}}}{2} + c$ , where $c$ is the constant of integration.

Note: To solve this type of question we need to simplify the given equation using logarithmic identities. A piece of good knowledge in formulas of integration of ${x^n}$ and differentiation of terms like $\log x$ and similar terms is appreciated. We should also be aware of some of the basic logarithmic properties. Writing the constant of integration at the end of the required integral is mandatory. Even if the integral is correct if the constant of integration is not present the answer is incomplete.