What is the integral of xe^{-x(1+y)} ?

The integral of xe^{-x(1+y)} is 1/((-1-y)^2)(-e^{-x(1+y)}x(1+y)-e^{-x(1+y)})+C

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integral of xe^{-x(1+y)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{( - 1 - y ) ^{2}} d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - (1 + y) x

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

Simplify

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

Add a constant to the solution

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

e^{2 t} t sin (5 t)

n = 0 \sum \infty (- 1)^{n} x^{2 n}

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

y = x^{2} - 3 x, y = 5 x

y^{^{'}} = 5 y + y^{5}, y (0) = 1

What is the integral of xe^{-x(1+y)} ?
The integral of xe^{-x(1+y)} is 1/((-1-y)^2)(-e^{-x(1+y)}x(1+y)-e^{-x(1+y)})+C