What is the integral of t^{11}e^{-t^6} ?

The integral of t^{11}e^{-t^6} is 1/6 (-e^{-t^6}t^6-e^{-t^6})+C

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integral of t^{11}e^{-t^6}

\int t^{11} e^{- t^{6}} d t

\frac{1}{6} (- e^{- t^{6}} t^{6} - e^{- t^{6}}) + C

Solution steps

\int t^{11} e^{- t^{6}} d t

Apply u-substitution

: \int \frac{e ^{u} u}{6} d u

= \int \frac{e ^{u} u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{6} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{6} (e^{u} u - e^{u})

Substitute back

u = - t^{6}

= \frac{1}{6} (e^{- t^{6}} (- t^{6}) - e^{- t^{6}})

Simplify

= \frac{1}{6} (- e^{- t^{6}} t^{6} - e^{- t^{6}})

Add a constant to the solution

= \frac{1}{6} (- e^{- t^{6}} t^{6} - e^{- t^{6}}) + C

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What is the integral of t^{11}e^{-t^6} ?
The integral of t^{11}e^{-t^6} is 1/6 (-e^{-t^6}t^6-e^{-t^6})+C