What is the integral of (8x^2)/(e^{2x)} ?

The integral of (8x^2)/(e^{2x)} is 8(-1/2 e^{-2x}x^2+1/4 (-2e^{-2x}x-e^{-2x}))+C

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integral of (8x^2)/(e^{2x)}

\int \frac{8 x ^{2}}{e ^{2 x}} d x

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

Solution steps

\int \frac{8 x ^{2}}{e ^{2 x}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2}}{e ^{2 x}} d x

Apply Integration By Parts

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

= 8 (- \frac{1}{2} e^{- 2 x} x^{2} - \int - e^{- 2 x} x d x)

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

= 8 (- \frac{1}{2} e^{- 2 x} x^{2} - (- \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x})))

Simplify

= 8 (- \frac{1}{2} e^{- 2 x} x^{2} + \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}))

Add a constant to the solution

= 8 (- \frac{1}{2} e^{- 2 x} x^{2} + \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x})) + C

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What is the integral of (8x^2)/(e^{2x)} ?
The integral of (8x^2)/(e^{2x)} is 8(-1/2 e^{-2x}x^2+1/4 (-2e^{-2x}x-e^{-2x}))+C