What is the integral of (6x^3+1)e^{3x^4+2x} ?

The integral of (6x^3+1)e^{3x^4+2x} is 1/2 e^{3x^4+2x}+C

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integral of (6x^3+1)e^{3x^4+2x}

\int (6 x^{3} + 1) e^{3 x^{4} + 2 x} d x

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

Solution steps

\int (6 x^{3} + 1) e^{3 x^{4} + 2 x} d x

Apply u-substitution

= \int \frac{e ^{u}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = 3 x^{4} + 2 x

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

Add a constant to the solution

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

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