What is the integral of t^3(e^t)^2 ?

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

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integral of t^3(e^t)^2

integral

t^{3} (e^{t})^{2}

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

Solution steps

\int t^{3} (e^{t})^{2} d t

(e^{t})^{2} = e^{2 t}

= \int e^{2 t} t^{3} d t

Apply Integration By Parts

: \frac{1}{2} e^{2 t} t^{3} - \int \frac{3}{2} e^{2 t} t^{2} d t

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

\int \frac{3}{2} e^{2 t} t^{2} d t = \frac{3}{2} (\frac{1}{2} e^{2 t} t^{2} - \frac{1}{4} (e^{2 t} \cdot 2 t - e^{2 t}))

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

Add a constant to the solution

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

∣ x - 2 ∣ 30

\frac{1}{x ^{6}}

2 e^{u}

e^{3 x^{2}}

x^{4} \cdot (e^{- x})^{2}

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