What is the integral of te^{3t}+e^t ?

The integral of te^{3t}+e^t is 1/9 (e^{3t}*3t-e^{3t})+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 te^{3t}+e^t

\int t e^{3 t} + e^{t} d t

\frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) + e^{t} + C

Solution steps

\int t e^{3 t} + e^{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int t e^{3 t} d t = \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t})

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

= \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) + e^{t}

Add a constant to the solution

= \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) + e^{t} + C

\int \frac{11 x ^{3}}{2 + x ^{2}} d x

\frac{d}{d x} ((1 + \frac{7}{x})^{x})

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

\frac{\partial}{\partial x} (ln (x) + 1)

\frac{d}{d x} (ln (x^{- 1}))

What is the integral of te^{3t}+e^t ?
The integral of te^{3t}+e^t is 1/9 (e^{3t}*3t-e^{3t})+e^t+C