What is the integral of t^{17}e^{-t^{18}} ?

The integral of t^{17}e^{-t^{18}} is -1/18 e^{-t^{18}}+C

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integral of t^{17}e^{-t^{18}}

\int t^{17} e^{- t^{18}} d t

- \frac{1}{18} e^{- t^{18}} + C

Solution steps

\int t^{17} e^{- t^{18}} d t

Apply u-substitution

: \int - \frac{e ^{u}}{18} d u

= \int - \frac{e ^{u}}{18} d u

Take the constant out:

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

= - \frac{1}{18} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{18} e^{u}

Substitute back

u = - t^{18}

= - \frac{1}{18} e^{- t^{18}}

Add a constant to the solution

= - \frac{1}{18} e^{- t^{18}} + C

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What is the integral of t^{17}e^{-t^{18}} ?
The integral of t^{17}e^{-t^{18}} is -1/18 e^{-t^{18}}+C