What is the integral of (1-t)e^{12t-6t^2} ?

The integral of (1-t)e^{12t-6t^2} is 1/12 e^{-6t^2+12t}+C

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integral of (1-t)e^{12t-6t^2}

\int (1 - t) e^{12 t - 6 t^{2}} d t

\frac{1}{12} e^{- 6 t^{2} + 12 t} + C

Solution steps

\int (1 - t) e^{12 t - 6 t^{2}} d t

Apply u-substitution

: \int - e^{- 6 u^{2} + 6} u d u

= \int - e^{- 6 u^{2} + 6} u d u

Take the constant out:

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

= - \int e^{- 6 u^{2} + 6} u d u

Apply u-substitution

: \int - \frac{e ^{v}}{12} d v

= - \int - \frac{e ^{v}}{12} d v

Take the constant out:

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

= - (- \frac{1}{12} \cdot \int e^{v} d v)

Use the common integral:

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

= - (- \frac{1}{12} e^{v})

Substitute back

= - (- \frac{1}{12} e^{- 6 (1 - t)^{2} + 6})

Simplify

- (- \frac{1}{12} e^{- 6 (1 - t)^{2} + 6}) : \frac{1}{12} e^{- 6 t^{2} + 12 t}

= \frac{1}{12} e^{- 6 t^{2} + 12 t}

Add a constant to the solution

= \frac{1}{12} e^{- 6 t^{2} + 12 t} + C

y^{^{'}} = y (x y^{2} + 2)

\int x (x + 1) d x

\frac{d}{d x} (\frac{x ^{2}}{1 + x ^{3}})

\int - \frac{1}{x ^{5}} d x

- \frac{x}{3 y}

What is the integral of (1-t)e^{12t-6t^2} ?
The integral of (1-t)e^{12t-6t^2} is 1/12 e^{-6t^2+12t}+C