Question

The integration of $ \int {\dfrac{{x\cos x\log x - \sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx $ is equal to
A. $ \dfrac{{\sin x}}{{\log x}} + c $
B. $ \log x + \sin x + c $
C. $ \log x\sin x + c $
D. $ \dfrac{{\cos x}}{{\log x}} + c $

Answer 1

Hint: To solve this problem, we will first separate the main term into two terms by separation of nominator. Then, we will use the integration by parts in the first term. After that, we will get our answer by simplifying the terms.
Formula used:
Integration by parts of two terms u and v is given by:
 \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx\]
Where, \[u\] is the function \[u\left( x \right)\] , \[v\] is the function \[v\left( x \right)\] and \[u'\] is the derivative of function \[u\left( x \right)\] .

Complete step-by-step answer:
We are asked to find the integration of $ \int {\dfrac{{x\cos x\log x - \sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx $
 $ I = \int {\dfrac{{x\cos x\log x - \sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx $
Now, we will separate the main term into two terms by separation of nominator.
 \[
   \Rightarrow I = \int {\dfrac{{x\cos x\log x}}{{x{{\left( {\log x} \right)}^2}}}} dx - \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx \\
   \Rightarrow I = \int {\dfrac{{\cos x}}{{\log x}}} dx - \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx \\
 \]
Now, we will consider the first term \[\int {\dfrac{{\cos x}}{{\log x}}} dx\]
We will use the integration by parts for solving this. Here, we will take $ u = \dfrac{1}{{\log x}} $ and $ v = \cos x $ and use the formula \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx\]
 \[
  \int {\dfrac{{\cos x}}{{\log x}}} dx \\
   = \dfrac{1}{{\log x}}\left( {\sin x} \right) + \int {\left( {\dfrac{{d\left( {\dfrac{1}{{\log x}}} \right)}}{{dx}} \cdot \int {\cos xdx} } \right)} dx \\
   = \dfrac{1}{{\log x}}\left( {\sin x} \right) + \int {\dfrac{1}{{x{{\left( {\log x} \right)}^2}}}} \sin xdx \\
   = \dfrac{{\sin x}}{{\log x}} + \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx \\
 \]
We will now put this in the main equation of integration.
 \[
   \Rightarrow I = \int {\dfrac{{x\cos x\log x}}{{x{{\left( {\log x} \right)}^2}}}} dx - \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx \\
   \Rightarrow I = \int {\dfrac{{\cos x}}{{\log x}}} dx - \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx \\
 \]
 \[ \Rightarrow I = \dfrac{{\sin x}}{{\log x}} + \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}dx} - \int {\dfrac{{\sin x}}{{x{{\left( {\log x} \right)}^2}}}dx} \]
 \[ \Rightarrow I = \dfrac{{\sin x}}{{\log x}} + c\]
Where, c is the constant of integration.
Thus, $ \int {\dfrac{{x\cos x\log x - \sin x}}{{x{{\left( {\log x} \right)}^2}}}} dx $ is equal to $ \dfrac{{\sin x}}{{\log x}} + c $ .
So, the correct answer is “Option A”.

Note: Here, we have used integration by parts method which is a special method of integration that is often useful when two functions are multiplied together. In this rule it is very important to choose write functions as u and v. There is a helpful rule to remember for choosing which function to take as u. This rule is called ILATE rule, where I stands for inverse trigonometric functions, L stands for logarithmic functions, A stands for algebraic functions, T stands for trigonometric functions and E stands for exponential functions.