Question

Find the value of \[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} \]
A) \[\log \left| {\dfrac{x}{{x + \cos x}}} \right| + c\]
B) \[\log \left| {\dfrac{{x + \cos x}}{x}} \right| + c\]
C) \[\log \left| {\dfrac{1}{{x + \cos x}}} \right| + c\]
D) \[\log \left| {x + \cos x} \right| + c\]

Answer 1

Hint:
We will rewrite the numerator of the integrand in such a way that there is a relation between the numerator and denominator. Then, we will separate the integrand and we will apply appropriate formulae of integration and find the value.
Formula used: \[\int {\dfrac{1}{x}dx = \log \left| x \right| + c} \], \[\log \left| a \right| - \log \left| b \right| = \log \left| {\dfrac{a}{b}} \right|\]

Complete step by step solution:
We have to integrate \[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} \]
Now we will try to make the numerator similar to the denominator.
Let us add and subtract \[x\] in the numerator. This gives us
\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} = \int {\dfrac{{\cos x + x\sin x + x - x}}{{x(x + \cos x)}}} dx\] ……….\[(1)\]
Let us club \[x\]and \[\cos x\] together. Also, we will club \[x\sin x\] and \[ - x\] together. Equation \[(1)\] becomes
\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} = \int {\dfrac{{(x + \cos x) + (x\sin x - x)}}{{x(x + \cos x)}}} dx\] ……….\[(2)\]
Now, we will separate the terms in the integrand as follows:
\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} = \int {\dfrac{{(x + \cos x)}}{{x(x + \cos x)}}} dx + \int {\dfrac{{(x\sin x - x)}}{{x(x + \cos x)}}dx} \] ……….\[(3)\]
Let us take \[{I_1} = \int {\dfrac{{x + \cos x}}{{x(x + \cos x)}}dx} \] and \[{I_2} = \int {\dfrac{{x\sin x - x}}{{x(x + \cos x)}}dx} \]
Now, in \[{I_1}\], the term \[x + \cos x\] is common in both the numerator and the denominator. So, we will cancel it out. Hence,
\[{I_1} = \int {\dfrac{1}{x}dx} = \log \left| x \right| + c\], where \[c\] is a constant
In \[{I_2}\], the factor \[x\] is common in the numerator. We will take it out. So,
\[{I_2} = \int {\dfrac{{x(\sin x - 1)}}{{x(x + \cos x)}}dx} \]
Both the numerator and denominator have \[x\] in common. So, we will cancel it out. This gives us
\[{I_2} = \int {\dfrac{{\sin x - 1}}{{x + \cos x}}dx} \]
We will use a substitution method to solve \[{I_2}\]. Let us take \[x + \cos x = u\]. Differentiating \[u\]with respect to \[x\], we get
\[\begin{array}{l}1 - \sin x = \dfrac{{du}}{{dx}}\\ \Rightarrow - (\sin x - 1)dx = du\\ \Rightarrow (\sin x - 1)dx = - du\end{array}\]
Rewriting \[{I_2}\] in terms of \[u\], we get
\[{I_2} = \int {\dfrac{{ - du}}{u} = - \int {\dfrac{1}{u}du} } \]
Therefore, we get \[{I_2}\] as
\[{I_2} = - \log \left| u \right| + c = - \log \left| {x + \cos x} \right| + c\]
Substituting the values of \[{I_1}\] and \[{I_2}\] in equation \[(3)\] we get
\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} = \log \left| x \right| - \log \left| {x + \cos x} \right| + c\]………..\[(4)\]
Using the property \[\log \left| a \right| - \log \left| b \right| = \log \left| {\dfrac{a}{b}} \right|\] in equation \[(4)\], we finally get
\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} = \log \left| {\dfrac{x}{{x + \cos x}}} \right| + c\]

Therefore, the correct option is A.

Note:
Here, we can observe that the denominator of the integrand is \[x(x + \cos x)\], which is a combination of algebraic and trigonometric functions. For such integrands, we will try to rewrite the numerator in such a way that there are similar terms in the numerator and the denominator. Integration is the inverse of differentiation and is used to find the summation of discrete data.