What is the integral of (6+x+9e^x) ?

The integral of (6+x+9e^x) is 6x+(x^2)/2+9e^x+C

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integral of (6+x+9e^x)

\int (6 + x + 9 e^{x}) d x

6 x + \frac{x ^{2}}{2} + 9 e^{x} + C

Solution steps

\int 6 + x + 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 6 d x + \int x d x + \int 9 e^{x} d x

\int 6 d x = 6 x

\int x d x = \frac{x ^{2}}{2}

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

= 6 x + \frac{x ^{2}}{2} + 9 e^{x}

Add a constant to the solution

= 6 x + \frac{x ^{2}}{2} + 9 e^{x} + C

\frac{d}{d x} (arccos (\frac{12}{x}))

\int \frac{sin ( x )}{1 + 2 cos ( x )} d x

\frac{d y}{d x} y = x^{- 8}

\int \frac{1}{( 1 - 6 t ) ^{4}} d t

cos^{2} (3 t)

What is the integral of (6+x+9e^x) ?
The integral of (6+x+9e^x) is 6x+(x^2)/2+9e^x+C