What is the integral of (2x-3)e^{x^2-3x+1} ?

The integral of (2x-3)e^{x^2-3x+1} is e^{x^2-3x+1}+C

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integral of (2x-3)e^{x^2-3x+1}

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

e^{x^{2} - 3 x + 1} + C

Solution steps

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

= e^{x^{2} - 3 x + 1}

Add a constant to the solution

= e^{x^{2} - 3 x + 1} + C

\int \frac{x ^{2} - x + 2}{x ^{4} + 10 x ^{2} + 9} d x

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = a

\frac{d}{d t} (e^{- 8 t})

\frac{5}{( s + 1 ) ( s + 2 ) ^{2}}

cos (3 \cdot x) \cdot cos (5 \cdot x)

What is the integral of (2x-3)e^{x^2-3x+1} ?
The integral of (2x-3)e^{x^2-3x+1} is e^{x^2-3x+1}+C