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

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

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

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

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

Solution steps

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

= e^{x^{3} - x}

Add a constant to the solution

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

35 x ln (10 x)

(1 + x^{4}) d y + x (1 + 4 y^{2}) d x = 0, y (1) = 0

4 y^{^{''}} - y = 1

\int e^{3 x} d x

(x^{2} - 9)^{2}

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