What is the integral of e^{x+1} ?

The integral of e^{x+1} is e^{1+x}+C

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integral of e^{x+1}

\int e^{x + 1} d x

e^{1 + x} + C

Solution steps

\int e^{x + 1} d x

e^{x + 1} = e^{1} e^{x}

= \int e^{1} e^{x} d x

Take the constant out:

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

= e^{1} \cdot \int e^{x} d x

Simplify

= e \cdot \int e^{x} d x

Use the common integral:

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

= e e^{x}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{x} = e^{1 + x}

= e^{1 + x}

Add a constant to the solution

= e^{1 + x} + C

\frac{d}{d x} (\frac{2}{x ^{2} + 4})

f (t) = t^{3} e^{- 2 t} cos (6 t)

\frac{\partial}{\partial x} (\frac{( x ^{2} + y ^{2} - 4 )}{x y})

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\frac{d}{d x} (4 x^{2} + 5)