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**Hint:**In this problem we have to calculate the integral value of the given equation. From this we are going to use the substitution method by substituting $u=x\cos x-\sin x-1$ and calculate the value of $du$. We will use the $uv$ formula of differentiation which is ${{\left( uv \right)}^{'}}=u{{v}^{'}}+v{{u}^{'}}$ and simplify the equation to get the value of $du$. After getting the value of $du$, we will substitute $u$, $du$ in the given integration and simplify the equation. Now we will use the integration formula $\int{\dfrac{dx}{x}}=\log \left| x \right|+C$ and calculate the required value.

**Complete step by step solution:**

Given that, $\int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=.....+C$.

Considering the integration part which is $\int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}$.

To solve the above integration part, we are going to use the substitution method. So, we are going to substitute $u=x\cos x-\sin x-1$ in the integration part. Before substituting this value, we also need the value of $du$. For this we are going to differentiate the value $u=x\cos x-\sin x-1$ with respect to $x$, then we will get

$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x-\sin x-1 \right)$

Applying the differentiation for each term, then we will get

$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x \right)-\dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}\left( 1 \right)$

We have the differentiation formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\dfrac{d}{dx}\left( 1 \right)=0$. Substituting these values in the above equation, then we will get

$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x \right)-\cos x$

Using the differentiation formula ${{\left( uv \right)}^{'}}=u{{v}^{'}}+v{{u}^{'}}$ in the above equation, then we will get

$\Rightarrow \dfrac{du}{dx}=x\dfrac{d}{dx}\left( \cos x \right)+\cos x\dfrac{d}{dx}\left( x \right)-\cos x$

We know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\left( x \right)=1$. Substituting these values and simplifying the equation, then we will get

$\begin{align}

& \Rightarrow \dfrac{du}{dx}=x\left( -\sin x \right)+\cos x-\cos x \\

& \Rightarrow du=-x\sin xdx \\

\end{align}$

From the values $u=x\cos x-\sin x-1$, $du=-x\sin xdx$, the given integration value modified as

$\begin{align}

& \Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=\int{\dfrac{1}{u}\left( -du \right)} \\

& \Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=-\int{\dfrac{du}{u}} \\

\end{align}$

We know that $\int{\dfrac{dx}{x}=\log \left| x \right|+C}$, then we will have

$\Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=-\log \left| x\cos x-\sin x-1 \right|+C$

**Hence option – C is the correct answer.**

**Note:**We can also use the formula $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\log \left| f\left( x \right) \right|+C$ for the above problem. We can use this formula after calculating the derivative of the denominator and use this formula to get the required result.

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