What is the integral of x^2e^{-10x} ?

The integral of x^2e^{-10x} is -1/500 e^{-10x}(50x^2+10x+1)+C

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integral of x^2e^{-10x}

\int x^{2} e^{- 10 x} d x

- \frac{1}{500} e^{- 10 x} (50 x^{2} + 10 x + 1) + C

Solution steps

\int x^{2} e^{- 10 x} d x

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{1000} d u

= \int - \frac{e ^{u} u ^{2}}{1000} d u

Take the constant out:

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

= - \frac{1}{1000} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{1000} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{1000} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 10 x

= - \frac{1}{1000} ((- 10 x)^{2} e^{- 10 x} - 2 (e^{- 10 x} (- 10 x) - e^{- 10 x}))

Simplify

- \frac{1}{1000} ((- 10 x)^{2} e^{- 10 x} - 2 (e^{- 10 x} (- 10 x) - e^{- 10 x})) : - \frac{1}{500} e^{- 10 x} (50 x^{2} + 10 x + 1)

= - \frac{1}{500} e^{- 10 x} (50 x^{2} + 10 x + 1)

Add a constant to the solution

= - \frac{1}{500} e^{- 10 x} (50 x^{2} + 10 x + 1) + C

\int - \frac{1}{x ^{3}} d x

x \to 5 + lim (I n (x^{2} - 25))

y^{^{''}} + 2 y^{^{'}} + y = t^{2} + 1 - e^{t}

\frac{d}{d x} (3 tan (x) + co tan (3 x))

\frac{d}{d x} (cos (2 x) y)

What is the integral of x^2e^{-10x} ?
The integral of x^2e^{-10x} is -1/500 e^{-10x}(50x^2+10x+1)+C