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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} - 1}{2} d u

= \int \frac{e ^{u} - 1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

= \frac{1}{2} (e^{u} - u)

Substitute back

u = x^{2}

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

Add a constant to the solution

= \frac{1}{2} (e^{x^{2}} - x^{2}) + C

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