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

\frac{d ^{2}}{d x ^{2}} (3 ln (sec (x) + tan (x)))

\int_{- 1}^{1} cos (x) d x

y^{^{''}} + 2 y = e^{x} + x^{3}

f (x) = sin^{6} (x)

\int \frac{e ^{x}}{1 - 4 e ^{2 x}} d 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