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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = x^{2} - 1

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

Add a constant to the solution

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

x \to 2 lim (\frac{11 x}{4 - x ^{2}})

(3 x - 5) (x^{2} + 4)

\int \frac{7 x + 14}{x ^{2} + 5 x - 6} d x

\int \frac{1}{3 x ^{2} - 2 x + 4} d x

y^{^{''}} + y = sec (t) tan (t)

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