What is the integral of 1/(1+\frac{x){2L}} ?

The integral of 1/(1+\frac{x){2L}} is 2Lln|1+x/(2L)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(1+\frac{x){2L}}

\int \frac{1}{1 + \frac{x}{2 L}} d x

2 L ln 1 + \frac{x}{2 L} + C

Solution steps

\int \frac{1}{1 + \frac{x}{2 L}} d x

Apply u-substitution

: \int \frac{2 L}{u} d u

= \int \frac{2 L}{u} d u

Take the constant out:

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

= 2 L \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 L ln ∣ u ∣

Substitute back

u = 1 + \frac{x}{2 L}

= 2 L ln 1 + \frac{x}{2 L}

Add a constant to the solution

= 2 L ln 1 + \frac{x}{2 L} + C

\int x^{2} e^{2 x^{3} + 1} d x

\int (x + 5) 2 - x d x

\int 2 x - 2 x^{3} d x

\int \frac{7 x ^{3}}{4 x ^{6}} d x

\int (2 - sin (\frac{x}{2}))^{2} d x

What is the integral of 1/(1+\frac{x){2L}} ?
The integral of 1/(1+\frac{x){2L}} is 2Lln|1+x/(2L)|+C