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

The integral of (3e^x+8x^3+4x) is 3e^x+2x^4+2x^2+C

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integral of (3e^x+8x^3+4x)

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

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

Solution steps

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x + x^{2} - 5))

x \to 0 - lim (h (x))

f (x) = ln (\frac{x ^{2}}{x ^{2} + 1})

\int_{1}^{6} \frac{4}{x ^{2} + 4 x + 4} d x

What is the integral of (3e^x+8x^3+4x) ?
The integral of (3e^x+8x^3+4x) is 3e^x+2x^4+2x^2+C