What is the integral of 4e^{x+y} ?

The integral of 4e^{x+y} is 4e^{y+x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4e^{x+y}

\int 4 e^{x + y}

4 e^{y + x} + C

Solution steps

\int 4 e^{x + y} d x

Take the constant out:

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

= 4 \cdot \int e^{y + x} d x

e^{y + x} = e^{y} e^{x}

= 4 \cdot \int e^{y} e^{x} d x

Take the constant out:

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

= 4 e^{y} \cdot \int e^{x} d x

Use the common integral:

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

= 4 e^{y} e^{x}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{y} e^{x} = e^{y + x}

= 4 e^{y + x}

Add a constant to the solution

= 4 e^{y + x} + C

\int cosh (u) d u

y = x^{2} - 5, (- 5, 20)

f (x) = \frac{1}{t ^{2}}

\int 43 tan^{5} (x) sec^{4} (x) d x

y = - x^{2} + 6 x, y = x^{2} - 4 x