What is the integral of e^{(-s+4)t}t ?

The integral of e^{(-s+4)t}t is 1/((4-s)^2)(e^{(4-s)t}(4-s)t-e^{(4-s)t})+C

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integral of e^{(-s+4)t}t

\int e^{(- s + 4) t} t d t

\frac{1}{( 4 - s ) ^{2}} (e^{(4 - s) t} (4 - s) t - e^{(4 - s) t}) + C

Solution steps

\int e^{(- s + 4) t} t d t

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{( 4 - s ) ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{( 4 - s ) ^{2}} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{( 4 - s ) ^{2}} (e^{u} u - e^{u})

Substitute back

u = (4 - s) t

= \frac{1}{( 4 - s ) ^{2}} (e^{(4 - s) t} (4 - s) t - e^{(4 - s) t})

Add a constant to the solution

= \frac{1}{( 4 - s ) ^{2}} (e^{(4 - s) t} (4 - s) t - e^{(4 - s) t}) + C

x \to 1 lim (\frac{x ^{- 2} - 6 ^{- 2}}{x + 6})

\int e^{(- 3 x)} d x

\int_{0}^{1} (x^{6} + x^{8}) d x

\frac{d y}{d x} + 4 x y = x e^{x^{2}}

x^{2} \cdot e^{x}

What is the integral of e^{(-s+4)t}t ?
The integral of e^{(-s+4)t}t is 1/((4-s)^2)(e^{(4-s)t}(4-s)t-e^{(4-s)t})+C