$\int \frac{x cos x}{{(x sin x + cos x)}^{2}} d x$

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Solution

$\int \frac{x cos x}{(x sin x + cos x)} = d x$

$p u t x sin x + cos x = u$

$d x = \frac{1}{x cos x} d u$

$\int \frac{d u}{u^{2}}$

$- \frac{1}{u} + c$

$\frac{1}{x sin x + cos x} + c$

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