What is the integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x} ?

The integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x} is 6x^{4/3}+1/2 e^{x^2+2x}+C

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integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x}

\int 8 3 x + (x + 1) e^{x^{2} + 2 x} d x

6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x} + C

Solution steps

\int 8 3 x + (x + 1) e^{x^{2} + 2 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 3 x d x + \int (x + 1) e^{x^{2} + 2 x} d x

\int 8 3 x d x = 6 x^{\frac{4}{3}}

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

= 6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x}

Add a constant to the solution

= 6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x} + C

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

\int cos (e^{t}) d t

\int_{0}^{5} (5 x - x^{2}) d x

\int 9 d x

x \to 3 lim (x^{3} - 3 x^{2} + 9 x)

What is the integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x} ?
The integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x} is 6x^{4/3}+1/2 e^{x^2+2x}+C