What is the integral of (w+rs-1/c e^{(r-p)t}) ?

The integral of (w+rs-1/c e^{(r-p)t}) is wt-e^{t(r-p)} 1/(c(r-p))+rst+C

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integral of (w+rs-1/c e^{(r-p)t})

\int (w + rs - \frac{1}{c} e^{(r - p) t}) d t

wt - e^{t (r - p)} \frac{1}{c ( r - p )} + rs t + C

Solution steps

\int w + rs - \frac{1}{c} e^{(r - p) t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int w d t - \int \frac{1}{c} e^{(r - p) t} d t + \int rs d t

\int w d t = wt

\int \frac{1}{c} e^{(r - p) t} d t = e^{t (r - p)} \frac{1}{c ( r - p )}

\int rs d t = rs t

= wt - e^{t (r - p)} \frac{1}{c ( r - p )} + rs t

Add a constant to the solution

= wt - e^{t (r - p)} \frac{1}{c ( r - p )} + rs t + C

\frac{d}{d x} ((sin (3 x))^{l n (x)})

n \to \infty lim (\frac{3}{2})

\frac{\partial}{\partial x} (\frac{x}{y})

\frac{d}{d x} (x^{2} 2 x)

y = ln (x^{100})

What is the integral of (w+rs-1/c e^{(r-p)t}) ?
The integral of (w+rs-1/c e^{(r-p)t}) is wt-e^{t(r-p)} 1/(c(r-p))+rst+C