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Question
Solution
\[\int\frac{\left( 1 + \cos x \right)}{\left( x + \sin x \right)^3}dx\]
\[\text{Let x} + \sin x = t\]
\[ \Rightarrow \left( 1 + \cos x \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 + \cos x \right) dx = dt\]
\[Now, \int\frac{\left( 1 + \cos x \right)}{\left( x + \sin x \right)^3}dx\]
\[ = \int\frac{dt}{t^3}\]
\[ = \int t^{- 3} dt\]
\[ = \frac{t^{- 3 + 1}}{- 3 + 1} + C\]
\[ = \frac{- 1}{2 t^2} + C\]
\[ = \frac{- 1}{\text{2} \left( x + \sin x \right)^2} + C\]
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