What is the integral of (x^2-1)e^{x^3-3x-5} ?

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

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integral of (x^2-1)e^{x^3-3x-5}

\int (x^{2} - 1) e^{x^{3} - 3 x - 5} d x

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

Solution steps

\int (x^{2} - 1) e^{x^{3} - 3 x - 5} d x

(x^{2} - 1) e^{x^{3} - 3 x - 5} = e^{- 5} (x^{2} - 1) e^{x^{3} - 3 x}

= \int e^{- 5} (x^{2} - 1) e^{x^{3} - 3 x} d x

Take the constant out:

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

= e^{- 5} \cdot \int (x^{2} - 1) e^{x^{3} - 3 x} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{5}} \cdot \int (x^{2} - 1) e^{x^{3} - 3 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{3} d u

= \frac{1}{e ^{5}} \cdot \int \frac{e ^{u}}{3} d u

Take the constant out:

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

= \frac{1}{e ^{5}} \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{e ^{5}} \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3} - 3 x

= \frac{1}{e ^{5}} \cdot \frac{1}{3} e^{x^{3} - 3 x}

Simplify

\frac{1}{e ^{5}} \cdot \frac{1}{3} e^{x^{3} - 3 x} : \frac{1}{3} e^{x^{3} - 3 x - 5}

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

Add a constant to the solution

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

f (x) = \frac{x ^{2}}{x + 4}, at x = 3

\frac{d}{d x} (e^{5 x^{4}})

\frac{d y}{d x} + \frac{y}{x} = 9 x + 10

x \to 4 lim (2 x - 5)

\int \frac{4}{x - 3} d x

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