What is the integral of e^{(2x^3-3x^2+5)}(x^2-x) ?

The integral of e^{(2x^3-3x^2+5)}(x^2-x) is 1/6 e^{2x^3-3x^2+5}+C

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integral of e^{(2x^3-3x^2+5)}(x^2-x)

\int e^{(2 x^{3} - 3 x^{2} + 5)} (x^{2} - x) d x

\frac{1}{6} e^{2 x^{3} - 3 x^{2} + 5} + C

Solution steps

\int e^{2 x^{3} - 3 x^{2} + 5} (x^{2} - x) d x

Apply u-substitution

= \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 2 x^{3} - 3 x^{2} + 5

= \frac{1}{6} e^{2 x^{3} - 3 x^{2} + 5}

Add a constant to the solution

= \frac{1}{6} e^{2 x^{3} - 3 x^{2} + 5} + C

What is the integral of e^{(2x^3-3x^2+5)}(x^2-x) ?
The integral of e^{(2x^3-3x^2+5)}(x^2-x) is 1/6 e^{2x^3-3x^2+5}+C