What is the integral of 9te^t ?

The integral of 9te^t is 9(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 9te^t

\int 9 t e^{t} d t

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

Solution steps

\int 9 t e^{t} d t

Take the constant out:

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

= 9 \cdot \int t e^{t} d t

Apply Integration By Parts

= 9 (e^{t} t - \int e^{t} d t)

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

= 9 (e^{t} t - e^{t})

Add a constant to the solution

= 9 (e^{t} t - e^{t}) + C

\int 15 x x + 1 d x

n = 2 \sum \infty \frac{1}{n ^{3} - n}

\int \frac{x ^{2} - 3 x - 1}{x ( x ^{2} + 4 )} d x

d er i v a t i v e x^{- 2}

\int 1 - x^{- 0.52} d x