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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{- 1 - y} e^{u}

Substitute back

u = - (1 + y) x

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

Add a constant to the solution

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

\int_{0}^{50} 9.8 (500 - 10 y) d y

\frac{\partial}{\partial t} (2 t)

f (x) = \frac{6}{4 x + 1}, at x = - 1

3 x^{5} - 3 x^{4} + x^{3}

y = \frac{( x ^{3} )}{3 ^{x}}

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