What is the integral of (t^2+1)e^{-2t} ?

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

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integral of (t^2+1)e^{-2t}

\int (t^{2} + 1) e^{- 2 t} d t

- \frac{1}{2} e^{- 2 t} t^{2} - \frac{1}{2} e^{- 2 t} t - \frac{3}{4} e^{- 2 t} + C

Solution steps

\int (t^{2} + 1) e^{- 2 t} d t

Apply Integration By Parts

: - \frac{1}{2} e^{- 2 t} (t^{2} + 1) - \int - e^{- 2 t} t d t

= - \frac{1}{2} e^{- 2 t} (t^{2} + 1) - \int - e^{- 2 t} t d t

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

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

Simplify

- \frac{1}{2} e^{- 2 t} (t^{2} + 1) - (- \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})) : - \frac{1}{2} e^{- 2 t} t^{2} - \frac{1}{2} e^{- 2 t} t - \frac{3}{4} e^{- 2 t}

= - \frac{1}{2} e^{- 2 t} t^{2} - \frac{1}{2} e^{- 2 t} t - \frac{3}{4} e^{- 2 t}

Add a constant to the solution

= - \frac{1}{2} e^{- 2 t} t^{2} - \frac{1}{2} e^{- 2 t} t - \frac{3}{4} e^{- 2 t} + C

y = 4 sin (x) cos (x), at x = \frac{π}{4}

4 x^{2} + 4 x + x y = 4

n \to \infty lim (1000 (1.06)^{n})

y^{^{''}} + y = 0, y (\frac{π}{3}) = 4, y^{^{'}} (\frac{π}{3}) = - 8

e^{x}, e^{- 4 x}, [0, ln (4)]

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