Question

What is the integration of $\int{\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)}dx}$ ?

Answer 1

Hint: Try to break the numerator into $\left( x+\cos x \right)-x\left( 1-\sin x \right)$ then break it. Apply integration separately and you’ll get the answer.

Complete step by step answer:

In the fraction $\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)}$ we will consider numerator first which is $\cos x+x\sin x$ which can be written as \[\left( x+\cos x-x+x\sin x \right)\] which can be further written as
\[\left( x+\cos x-\left( x-x\sin x \right) \right)\]
\[\Rightarrow \left( x+\cos x-x\left( 1-\sin x \right) \right)\]
So, \[\cos x+x\sin x\] can be written as \[\left( x+\cos x-x\left( 1-\sin x \right) \right)\]
Hence instead of fraction \[\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)}\] as \[\dfrac{\left( x+\cos x \right)-x\left( 1-\sin x \right)}{x\left( x+\cos x \right)}\]
Now, we will break it into two fractions which are
\[\dfrac{\left( x+\cos x \right)}{x\left( x+\cos x \right)}-\dfrac{x\left( 1-\sin x \right)}{x\left( x+\cos x \right)}\]
Hence, can be written as
\[\dfrac{1}{x}-\dfrac{1-\sin x}{x+\cos x}\]
Now we will consider this \[\dfrac{1}{x}-\dfrac{1-\sin x}{x+\cos x}\] to integrate.
Hence,
\[\int{\dfrac{\cos x+x\sin x}{x\left( x+cosx \right)}dx=}\int{\dfrac{1}{x}dx}-\int{\dfrac{1-\sin x}{x+\cos x}dx}\]
Now as we know that \[\int{\dfrac{1}{x}dx}=\ln \left( x \right)+{{c}_{1}}\]
So will write in next line and for \[\dfrac{1-\sin x}{x+\cos x}\] we will take \[x+\cos x=t\]
So, \[\left( 1-\sin x \right)dx=dt\]
Hence, in second line we write
\[=\ln \left( x \right)+{{c}_{1}}-\int{\dfrac{dt}{t}}\]
So as we know \[\int{\dfrac{dt}{t}}\] is \[\ln \left( t \right)+{{c}_{2}}\]
So,
\[\ln x+{{c}_{1}}-\int{\dfrac{dt}{t}}=\ln x+{{c}_{1}}-\left( \ln \left( t \right)+{{c}_{2}} \right)\]
\[=\ln x-\ln \left( t \right)+{{c}_{1}}-{{c}_{2}}\]
Hence, t was \[\left( x+\cos x \right)\] so we will substitute right back and instead \[{{c}_{1}}-{{c}_{2}}\] we can write simply c to represent constants.
$=\ln \left( x \right)-\ln \left( x+\cos x \right)+c$
Which can be further written as
$\ln \left( \dfrac{x}{x+\cos x} \right)+c$
Hence, the answer is $\ln \left( \dfrac{x}{x+\cos x} \right)+c$.

Note: Students should be careful while breaking the fractions and operating on them simultaneously. Also careful about calculations related to them.