What is the integral of (e^t+t^e) ?

The integral of (e^t+t^e) is e^t+(t^{e+1})/(e+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^t+t^e)

\int (e^{t} + t^{e}) d t

e^{t} + \frac{t ^{e + 1}}{e + 1} + C

Solution steps

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

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

\int t^{e} d t = \frac{t ^{e + 1}}{e + 1}

= e^{t} + \frac{t ^{e + 1}}{e + 1}

Add a constant to the solution

= e^{t} + \frac{t ^{e + 1}}{e + 1} + C

\frac{2}{x}, 8 x, \frac{x}{2}

\frac{\partial}{\partial x} (e^{x + x y - 2 y})

\int_{0}^{0.5} 600 x d x

f (x) = 5 x^{2} - 6 x + 6 + ln (x)

\frac{d ^{2}}{d x ^{2}} (4 y^{2} + 7 x^{2})

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