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

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

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integral of 1/(e^{-y)}

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

e^{y} + C

Solution steps

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

Simplify

\frac{1}{e ^{- y}} : e^{y}

= \int e^{y} d y

Use the common integral:

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

= e^{y}

Add a constant to the solution

= e^{y} + C

d y = 2 x (y^{2} + 9) d x

\int 0.6 x d x

x \to \infty lim (\frac{4}{7})

f (x) = x^{2} e^{2 x}

f (x) = \frac{3}{x + 2}