Question

Evaluate $I = \int {\dfrac{{dx}}{{4x + 5}}} {\text{ }}\left( {x \geqslant 0} \right)$
A. $4\ln \left( {4x + 5} \right)$
B. $4\ln \left( {4x + 5} \right) + C$
C. $\dfrac{{\ln \left( {4x + 5} \right)}}{4}$
D. $\dfrac{1}{4}\ln \left( {4x + 5} \right) + C$

Answer 1

In order to find the Integral of the function, first simplify the term by taking the denominator as a variable. Differentiate the obtained equation of the denominator with respect to $x$ , substitute the values in the main integral function. Solve the obtained equation using the formulas of integration known to us.
Formula used:
$\dfrac{{dx}}{{dx}} = 1$
$\dfrac{{d\left( {{\text{constant}}} \right)}}{{dx}} = 0$
$\int {\dfrac{1}{y}dy} = \ln y + C'$

Complete step-by-step answer :
We are given with an integral function $I = \int {\dfrac{{dx}}{{4x + 5}}}$ . ………(1)
Let us consider $4x + 5$ to be $y$ that is written as:
$y = 4x + 5$ ………(2)
Differentiating both the sides of equation (2), with respect to $x$ :
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {4x + 5} \right)}}{{dx}}$
Splitting the right side:
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {4x} \right)}}{{dx}} + \dfrac{{d\left( 5 \right)}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4dx}}{{dx}} + \dfrac{{d\left( 5 \right)}}{{dx}}$
Since, we know that $\dfrac{{dx}}{{dx}} = 1$ and $\dfrac{{d5}}{{dx}} = 0$ , so substituting these values in above equation, we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = 4 \times 1 + 0$
$\Rightarrow \dfrac{{dy}}{{dx}} = 4$
Multiplying both the sides by $dx$ , we get:
$\Rightarrow \dfrac{{dy}}{{dx}} \times dx = 4dx \\\ \Rightarrow dy = 4dx \;$
Dividing both the sides by $4$ , we get:
$\Rightarrow \dfrac{{dy}}{4} = \dfrac{{4dx}}{4} \\\ \Rightarrow \dfrac{{dy}}{4} = dx \\\ \Rightarrow dx = \dfrac{{dy}}{4} \;$
Substituting the value of $dx = \dfrac{{dy}}{4}$ and $y = 4x + 5$ in the equation (1):
$I = \int {\dfrac{{dx}}{{4x + 5}}} = \int {\dfrac{1}{y}\dfrac{{dy}}{4}}$
Taking the constant out of the integration, we get:
$I = \dfrac{1}{4}\int {\dfrac{1}{y}dy}$
From the formulas of integration, we know $\int {\dfrac{1}{y}dy} = \ln y + C'$ .
So, substituting this value in the above equation, we get:
$I = \dfrac{1}{4}\int {\dfrac{1}{y}dy}$
$\Rightarrow I = \dfrac{1}{4}\left( {\ln y + C'} \right)$
Opening the parenthesis:
$\Rightarrow I = \dfrac{1}{4}\ln y + \dfrac{1}{4}C'$
$\Rightarrow I = \dfrac{1}{4}\ln y + C$ where $C,C'$ are constants.
Substituting the value of $y$ from equation (2) in the above equation, and we get:
$\Rightarrow I = \dfrac{1}{4}\ln \left( {4x + 5} \right) + C$
Which matches with the Option D.
Therefore, $I = \int {\dfrac{{dx}}{{4x + 5}}} {\text{ = }}\dfrac{1}{4}\ln \left( {4x + 5} \right) + C$
Hence, Option D is correct.
So, the correct answer is “Option D”.

Note: Integration is nothing but the opposite of differentiation.
During differentiation, we know that the differentiation of a constant becomes zero, so the constant is removed, that’s why during integration a constant term is added to fulfil the requirement of that constant if removed.