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

The integral of e^{x+y^2-z} is e^{y^2-z+x}+C

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

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

e^{y^{2} - z + x} + C

Solution steps

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

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = y^{2} - z + x

= e^{y^{2} - z + x}

Add a constant to the solution

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

\int \frac{ln ( - 2 x )}{5 x} d x

t \to 0 lim (\frac{e ^{t} - 1}{t ^{9}})

y = x^{2} - 7, y = 6 x

6 y^{^{'''}} + 4 y^{^{''}} = 0

\int \frac{x}{e ^{x^{2}}} d x

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