What is the integral of e^{4t}-t^{-1} ?

The integral of e^{4t}-t^{-1} is 1/4 e^{4t}-ln|t|+C

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integral of e^{4t}-t^{-1}

\int e^{4 t} - t^{- 1} d t

\frac{1}{4} e^{4 t} - ln ∣ t ∣ + C

Solution steps

\int e^{4 t} - t^{- 1} 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 e^{4 t} d t - \int t^{- 1} d t

\int e^{4 t} d t = \frac{1}{4} e^{4 t}

\int t^{- 1} d t = ln ∣ t ∣

= \frac{1}{4} e^{4 t} - ln ∣ t ∣

Add a constant to the solution

= \frac{1}{4} e^{4 t} - ln ∣ t ∣ + C

\int - \frac{4}{x ( ln ( x ) ) ^{2}} d x

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\int \frac{x}{2 ( 1 + x )} d x

\int \frac{( y + 2 )}{y ^{2} + 4 y} d y

What is the integral of e^{4t}-t^{-1} ?
The integral of e^{4t}-t^{-1} is 1/4 e^{4t}-ln|t|+C