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

The integral of (2^{-3x^2})x is -(8^{-x^2})/(6ln(2))+C

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integral of (2^{-3x^2})x

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

- \frac{8 ^{- x^{2}}}{6 ln ( 2 )} + C

Solution steps

\int 2^{- 3 x^{2}} x d x

2^{- 3 x^{2}} = 8^{- x^{2}}

= \int 8^{- x^{2}} x d x

Apply u-substitution

: \int - \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

= \int - \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Take the constant out:

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

= - \frac{1}{3 ln ( 2 )} \cdot \int \frac{u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Apply Integration By Parts

: \frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u

= - \frac{1}{3 ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u)

\int - \frac{1}{4} d u = - \frac{1}{4} u

= - \frac{1}{3 ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u))

Substitute back

u = 8^{- x^{2}}

= - \frac{1}{3 ln ( 2 )} \frac{( 8 ^{- x^{2}} ) ^{2}}{2 ^{\frac{l n ( 2 \cdot 8 ^{- x^{2}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} \cdot 8^{- x^{2}})

Simplify

- \frac{1}{3 ln ( 2 )} \frac{( 8 ^{- x^{2}} ) ^{2}}{2 ^{\frac{l n ( 2 \cdot 8 ^{- x^{2}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} \cdot 8^{- x^{2}}) : - \frac{8 ^{- x^{2}}}{6 ln ( 2 )}

= - \frac{8 ^{- x^{2}}}{6 ln ( 2 )}

Add a constant to the solution

= - \frac{8 ^{- x^{2}}}{6 ln ( 2 )} + C

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What is the integral of (2^{-3x^2})x ?
The integral of (2^{-3x^2})x is -(8^{-x^2})/(6ln(2))+C