What is the integral of (4+x^{-4}-e^{3x+1}) ?

The integral of (4+x^{-4}-e^{3x+1}) is 4x-1/(3x^3)-1/3 e^{3x+1}+C

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integral of (4+x^{-4}-e^{3x+1})

\int (4 + x^{- 4} - e^{3 x + 1}) d x

4 x - \frac{1}{3 x ^{3}} - \frac{1}{3} e^{3 x + 1} + C

Solution steps

\int 4 + x^{- 4} - e^{3 x + 1} 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 4 d x + \int x^{- 4} d x - \int e^{3 x + 1} d x

\int 4 d x = 4 x

\int x^{- 4} d x = - \frac{1}{3 x ^{3}}

\int e^{3 x + 1} d x = \frac{1}{3} e^{3 x + 1}

= 4 x - \frac{1}{3 x ^{3}} - \frac{1}{3} e^{3 x + 1}

Add a constant to the solution

= 4 x - \frac{1}{3 x ^{3}} - \frac{1}{3} e^{3 x + 1} + C

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What is the integral of (4+x^{-4}-e^{3x+1}) ?
The integral of (4+x^{-4}-e^{3x+1}) is 4x-1/(3x^3)-1/3 e^{3x+1}+C