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

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

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

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

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

Solution steps

\int e^{3 x^{2} + 1} \cdot 4 x d x

Take the constant out:

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

= 4 \cdot \int e^{3 x^{2} + 1} x d x

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

= 4 \cdot \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{6} e^{u}

Substitute back

u = 3 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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