What is the integral of e^{5x}sqrt(e^{5x)+4} ?

The integral of e^{5x}sqrt(e^{5x)+4} is 2/15 (e^{5x}+4)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{5x}sqrt(e^{5x)+4}

\int e^{5 x} e^{5 x} + 4 d x

\frac{2}{15} (e^{5 x} + 4)^{\frac{3}{2}} + C

Solution steps

\int e^{5 x} e^{5 x} + 4 d x

Apply u-substitution

: \int \frac{u}{5} d u

= \int \frac{u}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = e^{5 x} + 4

= \frac{1}{5} \cdot \frac{2}{3} (e^{5 x} + 4)^{\frac{3}{2}}

Simplify

\frac{1}{5} \cdot \frac{2}{3} (e^{5 x} + 4)^{\frac{3}{2}} : \frac{2}{15} (e^{5 x} + 4)^{\frac{3}{2}}

= \frac{2}{15} (e^{5 x} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{15} (e^{5 x} + 4)^{\frac{3}{2}} + C

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

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

\frac{d}{d x} ((4 x - 1)^{3} (5 x + 3)^{2})

\int \frac{1}{( 1 + cos ^{2} ( x ) )} d x

What is the integral of e^{5x}sqrt(e^{5x)+4} ?
The integral of e^{5x}sqrt(e^{5x)+4} is 2/15 (e^{5x}+4)^{3/2}+C