Hi I need some help with this antiderivative.
cos(5x)+sinh(2x)
First, the antiderivative of a sum is the sum of the antiderivatives:
$\displaystyle \int \cos (5x) + \sinh (2x) dx = \int \cos (5x) dx + \int \sinh (2x) dx $
Then we look at $\displaystyle \cos (5x)$. The antiderivative of $\displaystyle \cos x$ is $\displaystyle \sin x$. By the chain rule
$\displaystyle \frac{d}{dx} \sin (5x) = \cos (5x) \cdot \frac{d}{dx} (5x) = 5 \cos (5x)$,
thus $\displaystyle \frac{d}{dx} (\tfrac{1}{5}\sin (5x)) = \tfrac{1}{5} \cdot 5 \cos (5x) = \cos (5x)$, and the antiderivative of $\displaystyle \cos(5x)$ is $\displaystyle \tfrac{1}{5}\sin (5x)$.
Similar reasoning from the antiderivative of $\displaystyle \sinh x$ will give you the other half of the antiderivative.
--Kevin C.