$\int \frac{(2 \sin x+5) \cos x}{\sin ^{2} x+5 \sin x} d x$

Answer

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Answer

$\boxed{2 ln(sin x + 5) + rac{5}{sin x + 5} + C}$

Steps

Step 1 ：Let $u = sin x + 5$, then $du = cos x dx$

Step 2 ：Substitute $u$ and $du$ into the integral: $int rac{(2u-5) du}{u^2} = int rac{2u-5}{u^2} du$

Step 3 ：Separate the integral: $int rac{2u}{u^2} du - int rac{5}{u^2} du$

Step 4 ：Simplify the integrals: $2 int rac{1}{u} du - 5 int rac{1}{u^2} du$

Step 5 ：Integrate: $2 ln|u| - 5 left(-rac{1}{u} ight) + C$

Step 6 ：Substitute back $u = sin x + 5$: $2 ln|sin x + 5| - 5 left(-rac{1}{sin x + 5} ight) + C$

Step 7 ：$\boxed{2 ln(sin x + 5) + rac{5}{sin x + 5} + C}$