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

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

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integral of e^{1-x}

integral

e^{1 - x}

- e^{1 - x} + C

Solution steps

\int e^{1 - x} d x

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = 1 - x

= - e^{1 - x}

Add a constant to the solution

= - e^{1 - x} + C

\frac{r ^{2}}{( a ^{2} - r ^{2} )}

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

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

\frac{( 6 x - 4 )}{( x ^{3} - 4 x )}

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