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

The integral of t^2e^{4t} is 1/4 e^{4t}t^2-1/32 (4e^{4t}t-e^{4t})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of t^2e^{4t}

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

\frac{1}{4} e^{4 t} t^{2} - \frac{1}{32} (4 e^{4 t} t - e^{4 t}) + C

Solution steps

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

Apply Integration By Parts

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

\int \frac{1}{2} e^{4 t} t d t = \frac{1}{32} (4 e^{4 t} t - e^{4 t})

= \frac{1}{4} e^{4 t} t^{2} - \frac{1}{32} (4 e^{4 t} t - e^{4 t})

Add a constant to the solution

= \frac{1}{4} e^{4 t} t^{2} - \frac{1}{32} (4 e^{4 t} t - e^{4 t}) + C

\int 9 x e^{6 x} d x

\frac{d y}{d x} = 4 + e^{y - 4 x + 5}

\frac{d}{d x} (e^{5 - 2 x})

x \to - \infty lim (- 8 x)

\int_{0}^{4} π (4 - y) d y

What is the integral of t^2e^{4t} ?
The integral of t^2e^{4t} is 1/4 e^{4t}t^2-1/32 (4e^{4t}t-e^{4t})+C