What is the integral of 3(x-1)y^3e^{y(1-x)} ?

The integral of 3(x-1)y^3e^{y(1-x)} is 3y(-ye^{-y(x-1)}(x-1)-e^{-y(x-1)})+C

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integral of 3(x-1)y^3e^{y(1-x)}

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

3 y (- y e^{- y (x - 1)} (x - 1) - e^{- y (x - 1)}) + C

Solution steps

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

Take the constant out:

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

= 3 y^{3} \cdot \int (- 1 + x) e^{(1 - x) y} d x

Apply u-substitution

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

= 3 y^{3} \cdot \int e^{- y u} u d u

Apply u-substitution

: \int \frac{e ^{v} v}{y ^{2}} d v

= 3 y^{3} \cdot \int \frac{e ^{v} v}{y ^{2}} d v

Take the constant out:

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

= 3 y^{3} \frac{1}{y ^{2}} \cdot \int e^{v} v d v

Apply Integration By Parts

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

= 3 y^{3} \frac{1}{y ^{2}} (e^{v} v - \int e^{v} d v)

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

= 3 y^{3} \frac{1}{y ^{2}} (e^{v} v - e^{v})

Substitute back

= 3 y^{3} \frac{1}{y ^{2}} (e^{- y (- 1 + x)} (- y (- 1 + x)) - e^{- y (- 1 + x)})

Simplify

3 y^{3} \frac{1}{y ^{2}} (e^{- y (- 1 + x)} (- y (- 1 + x)) - e^{- y (- 1 + x)}) : 3 y (- y e^{- y (x - 1)} (x - 1) - e^{- y (x - 1)})

= 3 y (- y e^{- y (x - 1)} (x - 1) - e^{- y (x - 1)})

Add a constant to the solution

= 3 y (- y e^{- y (x - 1)} (x - 1) - e^{- y (x - 1)}) + C

\int + e^{- 0.1 x} d x

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

\frac{1}{s ^{3} + 5 s}

\frac{1}{1 + s a}

\frac{d}{d x} f (x) = 8^{(x^{2} + 4 x)}

What is the integral of 3(x-1)y^3e^{y(1-x)} ?
The integral of 3(x-1)y^3e^{y(1-x)} is 3y(-ye^{-y(x-1)}(x-1)-e^{-y(x-1)})+C