What is the integral of-4x^2e^{-3x} ?

The integral of-4x^2e^{-3x} is 4/27 (9e^{-3x}x^2-2(-3e^{-3x}x-e^{-3x}))+C

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integral of-4x^2e^{-3x}

\int - 4 x^{2} e^{- 3 x} d x

\frac{4}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x})) + C

Solution steps

\int - 4 x^{2} e^{- 3 x} d x

Take the constant out:

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

= - 4 \cdot \int x^{2} e^{- 3 x} d x

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{27} d u

= - 4 \cdot \int - \frac{e ^{u} u ^{2}}{27} d u

Take the constant out:

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

= - 4 (- \frac{1}{27} \cdot \int e^{u} u^{2} d u)

Apply Integration By Parts

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

= - 4 (- \frac{1}{27} (u^{2} e^{u} - \int 2 u e^{u} d u))

Take the constant out:

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

= - 4 (- \frac{1}{27} (u^{2} e^{u} - 2 \cdot \int u e^{u} d u))

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= - 4 (- \frac{1}{27} (u^{2} e^{u} - 2 (e^{u} u - \int e^{u} d u)))

Use the common integral:

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

= - 4 (- \frac{1}{27} (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 3 x

= - 4 (- \frac{1}{27} ((- 3 x)^{2} e^{- 3 x} - 2 (e^{- 3 x} (- 3 x) - e^{- 3 x})))

Simplify

- 4 (- \frac{1}{27} ((- 3 x)^{2} e^{- 3 x} - 2 (e^{- 3 x} (- 3 x) - e^{- 3 x}))) : \frac{4}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x}))

= \frac{4}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x}))

Add a constant to the solution

= \frac{4}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x})) + C

x \to 1 lim ((f (x)) x^{2} - 2 x)

h \to 0 lim (h^{2} + 5^{h})

\int e^{- 7 x} d x

x \to \infty lim (1 + \frac{3}{x})

6 x^{2}, x^{2} + 2

What is the integral of-4x^2e^{-3x} ?
The integral of-4x^2e^{-3x} is 4/27 (9e^{-3x}x^2-2(-3e^{-3x}x-e^{-3x}))+C