What is the integral of x^7(4^{-x^8}) ?

The integral of x^7(4^{-x^8}) is -(4^{-x^8})/(16ln(2))+C

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integral of x^7(4^{-x^8})

\int x^{7} (4^{- x^{8}}) d x

- \frac{4 ^{- x^{8}}}{16 ln ( 2 )} + C

Solution steps

\int x^{7} \cdot 4^{- x^{8}} d x

Apply u-substitution

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

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

Apply Integration By Parts

: \frac{u ^{2}}{2 ^{\frac{l n ( u ) + 5 l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{32} d u

= - \frac{1}{ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( u ) + 5 l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{32} d u)

\int - \frac{1}{32} d u = - \frac{1}{32} u

= - \frac{1}{ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( u ) + 5 l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{32} u))

Substitute back

u = 4^{- x^{8}}

= - \frac{1}{ln ( 2 )} \frac{( 4 ^{- x^{8}} ) ^{2}}{2 ^{\frac{l n ( 4 ^{- x^{8}} ) + 5 l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{32} \cdot 4^{- x^{8}})

Simplify

- \frac{1}{ln ( 2 )} \frac{( 4 ^{- x^{8}} ) ^{2}}{2 ^{\frac{l n ( 4 ^{- x^{8}} ) + 5 l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{32} \cdot 4^{- x^{8}}) : - \frac{4 ^{- x^{8}}}{16 ln ( 2 )}

= - \frac{4 ^{- x^{8}}}{16 ln ( 2 )}

Add a constant to the solution

= - \frac{4 ^{- x^{8}}}{16 ln ( 2 )} + C

(- 2, 4), (3, a)

\frac{d}{d x} (50)

\int - 3 + r^{2} r d r

T + 2

y^{^{'}} = - y + 5

What is the integral of x^7(4^{-x^8}) ?
The integral of x^7(4^{-x^8}) is -(4^{-x^8})/(16ln(2))+C