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

Hint: 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.