What is the integral of 16te^t ?

The integral of 16te^t is 16(e^tt-e^t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 16te^t

\int 16 t e^{t} d t

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

Solution steps

\int 16 t e^{t} d t

Take the constant out:

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

= 16 \cdot \int t e^{t} d t

Apply Integration By Parts

: e^{t} t - \int e^{t} d t

= 16 (e^{t} t - \int e^{t} d t)

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

= 16 (e^{t} t - e^{t})

Add a constant to the solution

= 16 (e^{t} t - e^{t}) + C

\int (x^{3} - 4 x^{2} + 5) sin (4 x) d x

\int e^{x (a - s)} d x

\int \frac{5}{x ^{2}} d x

\int \frac{1}{4 x ^{2} - 16} d x

\int \frac{( x ^{4} + x ^{3} - 2 )}{( x ^{2} + 5 )} d x