What is the integral of x^2+9e^x+C ?

The integral of x^2+9e^x+C is (x^3)/3+9e^x+Cx+C

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integral of x^2+9e^x+C

\int x^{2} + 9 e^{x} + C d x

\frac{x ^{3}}{3} + 9 e^{x} + C x + C

Solution steps

\int x^{2} + 9 e^{x} + C 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 x^{2} d x + \int 9 e^{x} d x + \int C d x

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

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

\int C d x = C x

= \frac{x ^{3}}{3} + 9 e^{x} + C x

Add a constant to the solution

= \frac{x ^{3}}{3} + 9 e^{x} + C x + C

\frac{d}{d x} (1 + 5 sin (x))

\frac{d}{d x} (e^{a x} cos (b x))

\int_{1}^{2} \frac{( 3 x - 1 )}{3 x} d x

\frac{d}{d x} (sin (5 x) e^{3 x})

2 y^{^{'}} + 3 y = 0

What is the integral of x^2+9e^x+C ?
The integral of x^2+9e^x+C is (x^3)/3+9e^x+Cx+C