What is the integral of 18t^8e^{-t^9} ?

The integral of 18t^8e^{-t^9} is -2e^{-t^9}+C

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integral of 18t^8e^{-t^9}

\int 18 t^{8} e^{- t^{9}} d t

- 2 e^{- t^{9}} + C

Solution steps

\int 18 t^{8} e^{- t^{9}} d t

Take the constant out:

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

= 18 \cdot \int t^{8} e^{- t^{9}} d t

Apply u-substitution

: \int - \frac{e ^{u}}{9} d u

= 18 \cdot \int - \frac{e ^{u}}{9} d u

Take the constant out:

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

= 18 (- \frac{1}{9} \cdot \int e^{u} d u)

Use the common integral:

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

= 18 (- \frac{1}{9} e^{u})

Substitute back

u = - t^{9}

= 18 (- \frac{1}{9} e^{- t^{9}})

Simplify

18 (- \frac{1}{9} e^{- t^{9}}) : - 2 e^{- t^{9}}

= - 2 e^{- t^{9}}

Add a constant to the solution

= - 2 e^{- t^{9}} + C

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What is the integral of 18t^8e^{-t^9} ?
The integral of 18t^8e^{-t^9} is -2e^{-t^9}+C