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

The integral of t^8e^{-t^9} is -1/9 e^{-t^9}+C

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

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

- \frac{1}{9} e^{- t^{9}} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

= - \frac{1}{9} e^{u}

Substitute back

u = - t^{9}

= - \frac{1}{9} e^{- t^{9}}

Add a constant to the solution

= - \frac{1}{9} e^{- t^{9}} + C

y^{^{'}} + y = e^{(- t)}

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

\frac{d}{d x} (x - 3 sin (x))

\int \frac{e ^{u}}{( 4 - e ^{u} ) ^{2}} d u

\frac{d y}{d x} = y^{2} e^{- x}

What is the integral of t^8e^{-t^9} ?
The integral of t^8e^{-t^9} is -1/9 e^{-t^9}+C