What is the integral of 1/(x^2+1)-e^{-4x} ?

The integral of 1/(x^2+1)-e^{-4x} is arctan(x)+1/4 e^{-4x}+C

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integral of 1/(x^2+1)-e^{-4x}

\int \frac{1}{x ^{2} + 1} - e^{- 4 x} d x

arctan (x) + \frac{1}{4} e^{- 4 x} + C

Solution steps

\int \frac{1}{x ^{2} + 1} - e^{- 4 x} d x

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}{x ^{2} + 1} d x - \int e^{- 4 x} d x

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

\int e^{- 4 x} d x = - \frac{1}{4} e^{- 4 x}

= arctan (x) - (- \frac{1}{4} e^{- 4 x})

Simplify

= arctan (x) + \frac{1}{4} e^{- 4 x}

Add a constant to the solution

= arctan (x) + \frac{1}{4} e^{- 4 x} + C

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What is the integral of 1/(x^2+1)-e^{-4x} ?
The integral of 1/(x^2+1)-e^{-4x} is arctan(x)+1/4 e^{-4x}+C