What is the integral of 7e^{2x+e^{2x}} ?

The integral of 7e^{2x+e^{2x}} is 7/2 e^{e^{2x}}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7e^{2x+e^{2x}}

\int 7 e^{2 x + e^{2 x}} d x

\frac{7}{2} e^{e^{2 x}} + C

Solution steps

\int 7 e^{2 x + e^{2 x}} d x

Take the constant out:

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

= 7 \cdot \int e^{2 x + e^{2 x}} d x

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= 7 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} e^{u}

Substitute back

u = e^{2 x}

= 7 \cdot \frac{1}{2} e^{e^{2 x}}

Simplify

7 \cdot \frac{1}{2} e^{e^{2 x}} : \frac{7}{2} e^{e^{2 x}}

= \frac{7}{2} e^{e^{2 x}}

Add a constant to the solution

= \frac{7}{2} e^{e^{2 x}} + C

\frac{s ^{2}}{s ^{2} + 9}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 16 e ^{x} + 63} d x

ln (x^{6})

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

n = 0 \sum \infty 2 (- 5)^{n}

What is the integral of 7e^{2x+e^{2x}} ?
The integral of 7e^{2x+e^{2x}} is 7/2 e^{e^{2x}}+C