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

\frac{e ^{- s}}{s ^{2} + 3 s + 2}

\int \frac{e ^{- x}}{( 9 e ^{- 2 x} + 1 ) ^{\frac{3}{2}}} d x

1 - x^{2}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - y^{2})

\int sin (13 x) d x

What is the integral of (8x+1-9e^x) ?
The integral of (8x+1-9e^x) is 4x^2+x-9e^x+C