What is the integral of t^2e^{-t/4} ?

The integral of t^2e^{-t/4} is -4e^{-t/4}t^2+8(-4e^{-t/4}t-16e^{-t/4})+C

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integral of t^2e^{-t/4}

\int t^{2} e^{- \frac{t}{4}} d t

- 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}) + C

Solution steps

\int t^{2} e^{- \frac{t}{4}} d t

Apply Integration By Parts

= - 4 e^{- \frac{t}{4}} t^{2} - \int - 8 e^{- \frac{t}{4}} t d t

\int - 8 e^{- \frac{t}{4}} t d t = - 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}})

= - 4 e^{- \frac{t}{4}} t^{2} - (- 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}))

Simplify

= - 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}})

Add a constant to the solution

= - 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}) + C

What is the integral of t^2e^{-t/4} ?
The integral of t^2e^{-t/4} is -4e^{-t/4}t^2+8(-4e^{-t/4}t-16e^{-t/4})+C