What is the integral of 8x+1-9e^x ?

The integral of 8x+1-9e^x is 4x^2+x-9e^x+C

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integral of 8x+1-9e^x

\int 8 x + 1 - 9 e^{x} d x

4 x^{2} + x - 9 e^{x} + C

Solution steps

\int 8 x + 1 - 9 e^{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 8 x d x + \int 1 d x - \int 9 e^{x} d x

\int 8 x d x = 4 x^{2}

\int 1 d x = x

\int 9 e^{x} d x = 9 e^{x}

= 4 x^{2} + x - 9 e^{x}

Add a constant to the solution

= 4 x^{2} + x - 9 e^{x} + C

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