What is the integral of (2x-5)e^{x^2-5x} ?

The integral of (2x-5)e^{x^2-5x} is e^{x^2-5x}+C

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integral of (2x-5)e^{x^2-5x}

\int (2 x - 5) e^{x^{2} - 5 x} d x

e^{x^{2} - 5 x} + C

Solution steps

\int (2 x - 5) e^{x^{2} - 5 x} 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} - 5 x

= e^{x^{2} - 5 x}

Add a constant to the solution

= e^{x^{2} - 5 x} + C

\frac{d y}{d x} + 2 x y = 0

\int_{- 3}^{0} (2 + 9 - x^{2}) d x

n = 0 \sum \infty n e^{- 2 n}

x \to \infty lim (- \frac{10}{x})

y^{^{''}} + 16 y = - 32 sin (4 x)

What is the integral of (2x-5)e^{x^2-5x} ?
The integral of (2x-5)e^{x^2-5x} is e^{x^2-5x}+C