What is the integral of (3x^2-6x+9/x+4e^x) ?

The integral of (3x^2-6x+9/x+4e^x) is x^3-3x^2+9ln|x|+4e^x+C

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integral of (3x^2-6x+9/x+4e^x)

\int (3 x^{2} - 6 x + \frac{9}{x} + 4 e^{x}) d x

x^{3} - 3 x^{2} + 9 ln ∣ x ∣ + 4 e^{x} + C

Solution steps

\int 3 x^{2} - 6 x + \frac{9}{x} + 4 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 3 x^{2} d x - \int 6 x d x + \int \frac{9}{x} d x + \int 4 e^{x} d x

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

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

\int \frac{9}{x} d x = 9 ln ∣ x ∣

\int 4 e^{x} d x = 4 e^{x}

= x^{3} - 3 x^{2} + 9 ln ∣ x ∣ + 4 e^{x}

Add a constant to the solution

= x^{3} - 3 x^{2} + 9 ln ∣ x ∣ + 4 e^{x} + C

y^{^{'}} = 0.05 y - 12 \cdot 4000

\frac{d}{d x} (x sin (x + y) + y cos (x))

\frac{d}{d t} (\frac{t}{1 + t ^{2}})

x^{2}, 1, 0

f (x) = 9 x^{2} + 7 x - 6

What is the integral of (3x^2-6x+9/x+4e^x) ?
The integral of (3x^2-6x+9/x+4e^x) is x^3-3x^2+9ln|x|+4e^x+C