What is the integral of x^3(8^{-x^4}) ?

The integral of x^3(8^{-x^4}) is 16x^4+C

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integral of x^3(8^{-x^4})

\int x^{3} (8^{- x^{4}}) d x

16 x^{4} + C

Solution steps

\int x^{3} \cdot 8^{- x^{4}} d x

Apply u-substitution

= \int - 16 d u

Integral of a constant:

\int a d x = a x

= (- 16) u

Substitute back

u = - x^{4}

= (- 16) (- x^{4})

Simplify

= 16 x^{4}

Add a constant to the solution

= 16 x^{4} + C

\frac{d}{d x} (2 x^{3} - 11.7 x^{2} + 17.7 x - 5)

\int \frac{- 3 x ^{3}}{x ^{2} + 9} d x

\frac{d}{d x} (\frac{x y}{z})

\frac{\partial}{\partial x} ((x - y)^{5})

t an g e n t y = \frac{8}{x ^{2} + 4}, (2, 1)

What is the integral of x^3(8^{-x^4}) ?
The integral of x^3(8^{-x^4}) is 16x^4+C