What is the integral of (5/(e^{5t)}+1/(e^{6t)})/3 ?

The integral of (5/(e^{5t)}+1/(e^{6t)})/3 is 1/3 (-e^{-5t}-1/6 e^{-6t})+C

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integral of (5/(e^{5t)}+1/(e^{6t)})/3

\int \frac{\frac{5}{e ^{5 t}} + \frac{1}{e ^{6 t}}}{3} d t

\frac{1}{3} (- e^{- 5 t} - \frac{1}{6} e^{- 6 t}) + C

Solution steps

\int \frac{\frac{5}{e ^{5 t}} + \frac{1}{e ^{6 t}}}{3} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{5}{e ^{5 t}} + \frac{1}{e ^{6 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

= \frac{1}{3} (\int \frac{5}{e ^{5 t}} d t + \int \frac{1}{e ^{6 t}} d t)

\int \frac{5}{e ^{5 t}} d t = - e^{- 5 t}

\int \frac{1}{e ^{6 t}} d t = - \frac{1}{6} e^{- 6 t}

= \frac{1}{3} (- e^{- 5 t} - \frac{1}{6} e^{- 6 t})

Add a constant to the solution

= \frac{1}{3} (- e^{- 5 t} - \frac{1}{6} e^{- 6 t}) + C

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What is the integral of (5/(e^{5t)}+1/(e^{6t)})/3 ?
The integral of (5/(e^{5t)}+1/(e^{6t)})/3 is 1/3 (-e^{-5t}-1/6 e^{-6t})+C