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$$
\left(\int \frac{2 \cos x+1}{(2+\cos x)^2} d x\right)-\frac{\sin x}{2+\cos x}=
$$
Options:
\left(\int \frac{2 \cos x+1}{(2+\cos x)^2} d x\right)-\frac{\sin x}{2+\cos x}=
$$
Solution:
2267 Upvotes
Verified Answer
The correct answer is:
$\mathrm{C}$
Let $f(x)=\frac{\sin x}{2+\cos x}$
$$
\begin{aligned}
f^{\prime}(x) & =\frac{(2+\cos x) \cos x-\sin x(-\sin x)}{(2+\cos x)^2} \\
& =\frac{2 \cos x+1}{(2+\cos x)^2}
\end{aligned}
$$
$\frac{\sin x}{2+\cos x}$ is the antiderivatives of $\frac{2 \cos x+1}{(2+\cos x)^2}$
$$
\begin{aligned}
& \Rightarrow \int\left(\frac{2 \cos x+1}{(2+\cos x)^2}\right) d x=\frac{\sin x}{2+\cos x}+C \\
& \Rightarrow \int \frac{(2 \cos x+1)}{(2+\cos x)^2} d x-\frac{\sin x}{2+\cos x}=C
\end{aligned}
$$
$$
\begin{aligned}
f^{\prime}(x) & =\frac{(2+\cos x) \cos x-\sin x(-\sin x)}{(2+\cos x)^2} \\
& =\frac{2 \cos x+1}{(2+\cos x)^2}
\end{aligned}
$$
$\frac{\sin x}{2+\cos x}$ is the antiderivatives of $\frac{2 \cos x+1}{(2+\cos x)^2}$
$$
\begin{aligned}
& \Rightarrow \int\left(\frac{2 \cos x+1}{(2+\cos x)^2}\right) d x=\frac{\sin x}{2+\cos x}+C \\
& \Rightarrow \int \frac{(2 \cos x+1)}{(2+\cos x)^2} d x-\frac{\sin x}{2+\cos x}=C
\end{aligned}
$$
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