What is the integral of t^{15}e^{-t^8} ?

The integral of t^{15}e^{-t^8} is 1/8 (-e^{-t^8}t^8-e^{-t^8})+C

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integral of t^{15}e^{-t^8}

\int t^{15} e^{- t^{8}} d t

\frac{1}{8} (- e^{- t^{8}} t^{8} - e^{- t^{8}}) + C

Solution steps

\int t^{15} e^{- t^{8}} d t

Apply u-substitution

: \int \frac{e ^{u} u}{8} d u

= \int \frac{e ^{u} u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{8} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{8} (e^{u} u - e^{u})

Substitute back

u = - t^{8}

= \frac{1}{8} (e^{- t^{8}} (- t^{8}) - e^{- t^{8}})

Simplify

= \frac{1}{8} (- e^{- t^{8}} t^{8} - e^{- t^{8}})

Add a constant to the solution

= \frac{1}{8} (- e^{- t^{8}} t^{8} - e^{- t^{8}}) + C

\frac{d}{d x} (\frac{e ^{x} - cos ( x )}{ln ( x )})

\int_{0}^{1} x + 1 d x

\int \frac{6 x ^{3} + 5 x ^{2} - 46 x - 39}{x ^{2} - 9} d x

3 x d y - 2 y d x = 0

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

What is the integral of t^{15}e^{-t^8} ?
The integral of t^{15}e^{-t^8} is 1/8 (-e^{-t^8}t^8-e^{-t^8})+C