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

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

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

\int e^{3 x^{2} + l n (x)} d x

\frac{1}{6} e^{3 x^{2}} + C

Solution steps

\int e^{3 x^{2} + l n (x)} d x

Simplify

e^{3 x^{2} + l n (x)} : e^{3 x^{2}} x

= \int e^{3 x^{2}} x d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 3 x^{2}

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

Add a constant to the solution

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

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