What is the integral of x4^{-x^2} ?

The integral of x4^{-x^2} is -(4^{-x^2-1})/(ln(2))+C

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integral of x4^{-x^2}

\int x 4^{- x^{2}} d x

- \frac{4 ^{- x^{2} - 1}}{ln ( 2 )} + C

Solution steps

\int x \cdot 4^{- x^{2}} d x

Apply u-substitution

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

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

Apply Integration By Parts

: \frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{8} d u

= - \frac{1}{ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{8} d u)

\int - \frac{1}{8} d u = - \frac{1}{8} u

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

Substitute back

u = 4^{- x^{2}}

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

Simplify

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

= - \frac{4 ^{- x^{2} - 1}}{ln ( 2 )}

Add a constant to the solution

= - \frac{4 ^{- x^{2} - 1}}{ln ( 2 )} + C

\frac{1}{1 - x ^{2}}

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What is the integral of x4^{-x^2} ?
The integral of x4^{-x^2} is -(4^{-x^2-1})/(ln(2))+C