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

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.