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

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

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

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

e^{x^{2} - x} + C

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = x^{2} - x

= e^{x^{2} - x}

Add a constant to the solution

= e^{x^{2} - x} + C

f (t) = 1 \cdot t^{3}

\frac{d}{d x} (x + lo g_{e} (x))

x^{2} - x

\frac{d x}{d t} - 4 x = cos (5 t)

f (x) = - 3 x^{3} - 2 x^{2} - x + 3, (-, 5)

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