What is the integral of (1/(5-t)+60e^{-0.3t}+t) ?

The integral of (1/(5-t)+60e^{-0.3t}+t) is -ln|5-t|-199.99999…e^{-0.3t}+0.5t^2+C

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integral of (1/(5-t)+60e^{-0.3t}+t)

\int (\frac{1}{5 - t} + 60 e^{- 0.3 t} + t) d t

- ln ∣ 5 - t ∣ - 199.99999 \dots e^{- 0.3 t} + 0.5 t^{2} + C

Solution steps

\int \frac{1}{5 - t} + 60 e^{- 0.3 t} + 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 \frac{1}{5 - t} d t + \int 60 e^{- 0.3 t} d t + \int t d t

\int \frac{1}{5 - t} d t = - ln ∣ 5 - t ∣

\int 60 e^{- 0.3 t} d t = - 199.99999 \dots e^{- 0.3 t}

\int t d t = \frac{t ^{2}}{2}

= - ln ∣ 5 - t ∣ - 199.99999 \dots e^{- 0.3 t} + \frac{t ^{2}}{2}

Simplify

= - ln ∣ 5 - t ∣ - 199.99999 \dots e^{- 0.3 t} + 0.5 t^{2}

Add a constant to the solution

= - ln ∣ 5 - t ∣ - 199.99999 \dots e^{- 0.3 t} + 0.5 t^{2} + C

\int 3 x ln (5 x) d x

\int \frac{14}{sec ( 2 x ) - 1} d x

f (x) = 5 x^{4} + 8 sin (x)

(- e^{- x})^{^{'}}

\frac{s}{s ^{2} + 4}

What is the integral of (1/(5-t)+60e^{-0.3t}+t) ?
The integral of (1/(5-t)+60e^{-0.3t}+t) is -ln|5-t|-199.99999…e^{-0.3t}+0.5t^2+C