What is the integral of 4x^3+e^x-1 ?

The integral of 4x^3+e^x-1 is x^4+e^x-x+C

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integral of 4x^3+e^x-1

\int 4 x^{3} + e^{x} - 1 d x

x^{4} + e^{x} - x + C

Solution steps

\int 4 x^{3} + e^{x} - 1 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 4 x^{3} d x + \int e^{x} d x - \int 1 d x

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

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

\int 1 d x = x

= x^{4} + e^{x} - x

Add a constant to the solution

= x^{4} + e^{x} - x + C

x \to 2 lim (\frac{x + 2}{x ^{3} + 8})

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

\frac{x ^{5} - 5}{x + 1}

x \to 4 lim (9 x + ∣ x - 4 ∣)