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

The integral of x^2e^{-2ax} is -(e^{-2ax}(2a^2x^2+2ax+1))/(4a^3)+C

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

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

- \frac{e ^{- 2 a x} ( 2 a ^{2} x ^{2} + 2 a x + 1 )}{4 a ^{3}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{ln ^{2} ( u )}{8 a ^{3}} d u

= \int - \frac{ln ^{2} ( u )}{8 a ^{3}} d u

Take the constant out:

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

= - \frac{1}{8 a ^{3}} \cdot \int ln^{2} (u) d u

Apply Integration By Parts

: u ln^{2} (u) - \int 2 ln (u) d u

= - \frac{1}{8 a ^{3}} (u ln^{2} (u) - \int 2 ln (u) d u)

\int 2 ln (u) d u = 2 (u ln (u) - u)

= - \frac{1}{8 a ^{3}} (u ln^{2} (u) - 2 (u ln (u) - u))

Substitute back

u = e^{- 2 a x}

= - \frac{1}{8 a ^{3}} (e^{- 2 a x} ln^{2} (e^{- 2 a x}) - 2 (e^{- 2 a x} ln (e^{- 2 a x}) - e^{- 2 a x}))

Simplify

- \frac{1}{8 a ^{3}} (e^{- 2 a x} ln^{2} (e^{- 2 a x}) - 2 (e^{- 2 a x} ln (e^{- 2 a x}) - e^{- 2 a x})) : - \frac{e ^{- 2 a x} ( 2 a ^{2} x ^{2} + 2 a x + 1 )}{4 a ^{3}}

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

Add a constant to the solution

= - \frac{e ^{- 2 a x} ( 2 a ^{2} x ^{2} + 2 a x + 1 )}{4 a ^{3}} + C

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What is the integral of x^2e^{-2ax} ?
The integral of x^2e^{-2ax} is -(e^{-2ax}(2a^2x^2+2ax+1))/(4a^3)+C