What is the integral of x^{15}e^{-x^{16}} ?

The integral of x^{15}e^{-x^{16}} is -1/16 e^{-x^{16}}+C

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integral of x^{15}e^{-x^{16}}

\int x^{15} e^{- x^{16}} d x

- \frac{1}{16} e^{- x^{16}} + C

Solution steps

\int x^{15} e^{- x^{16}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{16} d u

= \int - \frac{e ^{u}}{16} d u

Take the constant out:

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

= - \frac{1}{16} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{16} e^{u}

Substitute back

u = - x^{16}

= - \frac{1}{16} e^{- x^{16}}

Add a constant to the solution

= - \frac{1}{16} e^{- x^{16}} + C

f (t) = 2 t^{3} - 3 t^{2} - 4 t

\frac{\partial}{\partial x} (\frac{4 π ^{2} x}{y ^{2}})

9 x e^{x} csc (x)

\frac{1}{x}, 1, 2

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

What is the integral of x^{15}e^{-x^{16}} ?
The integral of x^{15}e^{-x^{16}} is -1/16 e^{-x^{16}}+C