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

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

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integral of ((x-1)e^{x^2})/(e^{2x)}

\int \frac{( x - 1 ) e ^{x^{2}}}{e ^{2 x}} d x

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

Solution steps

\int \frac{( x - 1 ) e ^{x^{2}}}{e ^{2 x}} d x

Simplify

x^{2} : xx

= \int \frac{( x - 1 ) e ^{xx}}{e ^{2 x}} d x

Simplify

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

= \int e^{x^{2} - 2 x} (x - 1) d x

Apply u-substitution

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

= \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

= \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = x^{2} - 2 x

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

Add a constant to the solution

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

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