What is the integral of (e^{-z})/((x^2+y^2)) ?

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

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integral of (e^{-z})/((x^2+y^2))

integral

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

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

Solution steps

\int \frac{e ^{- z}}{x ^{2} + y ^{2}} d z

Take the constant out:

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

= \frac{1}{x ^{2} + y ^{2}} \cdot \int e^{- z} d z

Apply u-substitution

: \int - e^{u} d u

= \frac{1}{x ^{2} + y ^{2}} \cdot \int - e^{u} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - z

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

Simplify

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

Add a constant to the solution

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

\frac{1}{2 x - 1}

1 + t^{2}

\frac{( x ^{2} + 2 x + 1 )}{x ^{2}}

\frac{2}{( x ^{2} + 1 ) ^{2}}

\frac{1}{( x ^{2} + 2 ^{2} )}

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