What is the integral of (8x^3-9/x+4e^x) ?

The integral of (8x^3-9/x+4e^x) is 2x^4-9ln|x|+4e^x+C

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integral of (8x^3-9/x+4e^x)

\int (8 x^{3} - \frac{9}{x} + 4 e^{x}) d x

2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x} + C

Solution steps

\int 8 x^{3} - \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 8 x^{3} d x - \int \frac{9}{x} d x + \int 4 e^{x} d x

\int 8 x^{3} d x = 2 x^{4}

\int \frac{9}{x} d x = 9 ln ∣ x ∣

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

= 2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x}

Add a constant to the solution

= 2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x} + C

What is the integral of (8x^3-9/x+4e^x) ?
The integral of (8x^3-9/x+4e^x) is 2x^4-9ln|x|+4e^x+C