What is the integral of t^7e^{-t^4} ?

The integral of t^7e^{-t^4} is 1/4 (-e^{-t^4}t^4-e^{-t^4})+C

integral of t^7e^{-t^4}

\int t^{7} e^{- t^{4}} d t

\frac{1}{4} (- e^{- t^{4}} t^{4} - e^{- t^{4}}) + C

Solution steps

\int t^{7} e^{- t^{4}} d t

Apply u-substitution

= \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = - t^{4}

= \frac{1}{4} (e^{- t^{4}} (- t^{4}) - e^{- t^{4}})

Simplify

= \frac{1}{4} (- e^{- t^{4}} t^{4} - e^{- t^{4}})

Add a constant to the solution

= \frac{1}{4} (- e^{- t^{4}} t^{4} - e^{- t^{4}}) + C

\frac{d}{d x} (\frac{3 x ^{4} + 3}{sin ( x )})

\frac{d}{d s} (s^{2} + t^{2})

t an g e n t 3 x^{4} - 7 x^{2}, at x = - 1

\int \frac{- 2 x + 4}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

x \to - 3 lim (\frac{1}{2} - x^{3})

What is the integral of t^7e^{-t^4} ?
The integral of t^7e^{-t^4} is 1/4 (-e^{-t^4}t^4-e^{-t^4})+C