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

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

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

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

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

Solution steps

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

Simplify

= \int e^{- 3 x^{2} - 1} x d x

Apply u-substitution

= \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} - 1

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

Add a constant to the solution

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

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