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

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

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integral of 1/(e^{x+1)}

\int \frac{1}{e ^{x + 1}} d x

- \frac{1}{e ^{x + 1}} + C

Solution steps

\int \frac{1}{e ^{x + 1}} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

= - \frac{1}{e ^{x + 1}}

Add a constant to the solution

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

\int (x^{3} + 1)^{2} d x

\frac{( x ^{2} - 2 x + 3 )}{( x - 1 ) ^{2}}

\int_{2 y}^{2} \int_{0}^{1} e^{x^{2}} d y d x

\frac{d}{d x} (e^{3 x} sin (2 x))

y d x + x^{2} d y = 0

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