What is the integral of 4e^x+12x^3+4x ?

The integral of 4e^x+12x^3+4x is 4e^x+3x^4+2x^2+C

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integral of 4e^x+12x^3+4x

\int 4 e^{x} + 12 x^{3} + 4 x d x

4 e^{x} + 3 x^{4} + 2 x^{2} + C

Solution steps

\int 4 e^{x} + 12 x^{3} + 4 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 4 e^{x} d x + \int 12 x^{3} d x + \int 4 x d x

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

\int 12 x^{3} d x = 3 x^{4}

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

= 4 e^{x} + 3 x^{4} + 2 x^{2}

Add a constant to the solution

= 4 e^{x} + 3 x^{4} + 2 x^{2} + C

f (x) = 2 sin (x), at x = \frac{π}{4}

(5.2) (- 9.6)

e^{x^{2} - x}

y = 3 + 4 x^{2} - 2 x^{3} x = a

x \to - \infty lim (- 3 x)

What is the integral of 4e^x+12x^3+4x ?
The integral of 4e^x+12x^3+4x is 4e^x+3x^4+2x^2+C