What is the integral of x/(e^{x^2-8)} ?

The integral of x/(e^{x^2-8)} is -1/2 e^{8-x^2}+C

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integral of x/(e^{x^2-8)}

\int \frac{x}{e ^{x^{2} - 8}} d x

- \frac{1}{2} e^{8 - x^{2}} + C

Solution steps

\int \frac{x}{e ^{x^{2} - 8}} d x

\frac{x}{e ^{x^{2} - 8}} = \frac{1}{e ^{- 8}} \frac{x}{e ^{x^{2}}}

= \int \frac{1}{e ^{- 8}} \cdot \frac{x}{e ^{x^{2}}} d x

Take the constant out:

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

= \frac{1}{e ^{- 8}} \cdot \int \frac{x}{e ^{x^{2}}} d x

\frac{1}{e ^{- 8}} = e^{8}

= e^{8} \cdot \int \frac{x}{e ^{x^{2}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x}{e ^{x^{2}}} = (x) e^{- x^{2}}

= e^{8} \cdot \int x e^{- x^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{2} d u

= e^{8} \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= e^{8} (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= e^{8} (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= e^{8} (- \frac{1}{2} e^{- x^{2}})

Simplify

e^{8} (- \frac{1}{2} e^{- x^{2}}) : - \frac{1}{2} e^{8 - x^{2}}

= - \frac{1}{2} e^{8 - x^{2}}

Add a constant to the solution

= - \frac{1}{2} e^{8 - x^{2}} + C

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What is the integral of x/(e^{x^2-8)} ?
The integral of x/(e^{x^2-8)} is -1/2 e^{8-x^2}+C