What is the integral of (e^x)/(sqrt(e^{2x)-e^x)} ?

The integral of (e^x)/(sqrt(e^{2x)-e^x)} is 2ln|e^{x/2}+sqrt(e^x-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^x)/(sqrt(e^{2x)-e^x)}

\int \frac{e ^{x}}{e ^{2 x} - e ^{x}} d x

2 ln e^{\frac{x}{2}} + e^{x} - 1 + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} - e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - u} d u

= \int \frac{1}{u ^{2} - u} d u

Factor

\frac{1}{u ^{2} - u}

= \int \frac{1}{u ( u - 1 )} d u

Apply u-substitution

: \int \frac{1}{( v + 1 ) v} d v

= \int \frac{1}{( v + 1 ) v} d v

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(v + 1) v = (v + 1) v

= \int \frac{1}{v + 1 v} d v

Apply u-substitution

: \int \frac{2}{w ^{2} + 1} d w

= \int \frac{2}{w ^{2} + 1} d w

Take the constant out:

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

= 2 \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = ln w^{2} + 1 + w

= 2 ln w^{2} + 1 + w

Substitute back

= 2 ln (e^{x} - 1)^{2} + 1 + e^{x} - 1

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

= 2 ln e^{\frac{x}{2}} + e^{x} - 1

Add a constant to the solution

= 2 ln e^{\frac{x}{2}} + e^{x} - 1 + C

\int \frac{3 cos ( ln ∣ x ∣ )}{x} d x

x \to 0 lim (x tan (x))

\frac{x ^{6} + 3 x ^{2} + 50}{x ^{2} + 1}

f (x) = \frac{2 x}{( x ^{3} - 9 ) ^{2}}

\frac{d}{d x} (cot (6 x))

What is the integral of (e^x)/(sqrt(e^{2x)-e^x)} ?
The integral of (e^x)/(sqrt(e^{2x)-e^x)} is 2ln|e^{x/2}+sqrt(e^x-1)|+C