$\int \frac{(1 + x) \sin x}{(x^{2} + 2 x) \cos^{2} x - (1 + x) \sin 2 x} dx$

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$\frac{1}{2} \log_{e} |\frac{\sin x - (x + 1) \cos x - 1}{\sin x - (x + 1) \cos x + 1}| + C$

$\frac{1}{2} \tan^{- 1} {\sin x - (x + 1) \cos x} + C$

$\frac{1}{2} \sin^{- 1} {\sin x - (x + 1) \cos x} + C$

$\frac{1}{2} \sin^{- 1} (\cos x + \sin x) + C$

answer is A.

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Detailed Solution

$\begin{aligned} I = \int \frac{(1 + x) \sin xdx}{(x + 1)^{2} \cos^{2} x - \cos^{2} x - 2 (1 + x) \sin xcos x} \\ I = \frac{(1 + x) \sin xdx}{(x + 1)^{2} \cos^{2} x + \sin^{2} x - 2 (1 + x) \sin xcos x - 1} \end{aligned}$

$I = \frac{(1 + x) \sin x}{(\sin x - (x + 1) \cos x)^{2} - 1} dx$

$Put \sin x - (x + 1) \cos x = t \Rightarrow \sin x (x + 1) dx = dt$

$\begin{array}{l} ∴ I = \int \frac{dt}{t^{2} - 1} = \frac{1}{2} \log |\frac{t - 1}{t + 1}| + C \\ = \frac{1}{2} \log_{e} |\frac{\sin x - (x + 1) \cos x - 1}{\sin x - (x + 1) \cos x + 1}| + C \end{array}$