What is the integral of te^{t+2} ?

The integral of te^{t+2} is e^2(e^tt-e^t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of te^{t+2}

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

e^{2} (e^{t} t - e^{t}) + C

Solution steps

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

t e^{t + 2} = e^{2} t e^{t}

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Add a constant to the solution

= e^{2} (e^{t} t - e^{t}) + C

\int \frac{d m}{m ^{3} + m ^{2}} d x

\frac{d}{d r} (\frac{1}{2 r} + \frac{1}{7 r ^{\frac{6}{7}}})

\int cos (t) sin^{4} (t) d t

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

\int 21 x^{20} d x