Question

\[\int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} \] is equal to
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: This is an integration question. To solve this question first we will add and subtract x from the numerator, then we will separate and integrate them. We will convert the complex term into simpler taking it in terms of another variable and integrate that and then again will convert the answer to original form.

Complete step-by-step answer:
Let’s take \[I = \int {\dfrac{{\cos x + x\sin x}}{{x(x + \cos x)}}dx} \]
Adding a x and subtracting a x in numerator we get,
\[I = \int {\dfrac{{\cos x + x + x\sin x - x}}{{x(x + \cos x)}}dx} \]
Taking $ x + \cos x $ of the numerator into a bracket and taking x common from $ x\sin x - x $ we get,
\[I = \int {\dfrac{{(\cos x + x) + x(\sin x - 1)}}{{x(x + \cos x)}}dx} \]
Separating the terms of numerator with same denominator we get
\[I = \int {\left( {\dfrac{{(\cos x + x)}}{{x(x + \cos x)}} + \dfrac{{x(\sin x - 1)}}{{x(x + \cos x)}}} \right)dx} \]
Taking integration of separated terms separately we get,
\[I = \int {\dfrac{{(\cos x + x)}}{{x(x + \cos x)}}dx + \int {\dfrac{{x(\sin x - 1)}}{{x(x + \cos x)}}} dx} \]
Cancelling ( $ x + \cos x $ ) and x from numerator and denominator in respective separated terms of right hand side we get,
\[I = \int {\dfrac{1}{x}dx + \int {\dfrac{{(\sin x - 1)}}{{(x + \cos x)}}} dx} \] ……………. (1)
Now let’s take, $ x + \cos x = t $
Differentiating the above equation we get,
 $ (1 - \sin x)dx = dt $
Taking -1 common from left hand side we get,
 $ - (\sin x - 1)dx = dt $
Again multiplying -1 on both side we get,
 $ (\sin x - 1)dx = - dt $
Putting data obtained from above equations in equation 1 we get,
\[I = \int {\dfrac{1}{x}dx + \int {\dfrac{{ - dt}}{t}} } \]
Taking -1 out and changing + sign into – sign we get,
\[I = \int {\dfrac{1}{x}dx - \int {\dfrac{1}{t}dt} } \]
As we know the formula of integration $ \int {\dfrac{1}{x}dx = \ln x = \log \left| x \right|} $
Substituting the above formula in the above equation we get,
\[I = \log \left| x \right| - \log \left| t \right| + C\]
C is the integration constant.
As we know the formula of log that $ \log a - \log b = \log \dfrac{a}{b} $
Substituting above formula in the above equation we get,
\[I = \log \left| {\dfrac{x}{{x + \cos x}}} \right| + C\]

Option A is correct.

Note: Integration is the technique of finding a function g(x) from its derivative dg(x), which is equal to a given function f(x).
The derivative of an integral of a function is that original function or we can say differentiation undoes the result of integration.
To solve this type of question you should remember all the formulas, properties and rules of trigonometry, inverse trigonometry, derivative and integration and logarithm.
You might get confused in equation 1 for getting complex functions. In this type of case, try to convert that function into simpler form in another variable form.