What is the integral of (e^x-6x)/4 ?

The integral of (e^x-6x)/4 is 1/4 (e^x-3x^2)+C

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integral of (e^x-6x)/4

\int \frac{e ^{x} - 6 x}{4} d x

\frac{1}{4} (e^{x} - 3 x^{2}) + C

Solution steps

\int \frac{e ^{x} - 6 x}{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int e^{x} - 6 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

= \frac{1}{4} (\int e^{x} d x - \int 6 x d x)

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

\int 6 x d x = 3 x^{2}

= \frac{1}{4} (e^{x} - 3 x^{2})

Add a constant to the solution

= \frac{1}{4} (e^{x} - 3 x^{2}) + C

What is the integral of (e^x-6x)/4 ?
The integral of (e^x-6x)/4 is 1/4 (e^x-3x^2)+C