You need to use paretheses. Is this what you mean?

$\dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6$

or did you mean:

$\dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}$

Assuming it is the latter:

Let $u = \cos(5x)+5x+6$. Then $du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx$

This gives:

$\displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C$

Plugging in the initial values:

$3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)$

This gives $y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3$