What is the integral of (2x)/(x^2-1) ?

The integral of (2x)/(x^2-1) is ln|x^2-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(x^2-1)

\int \frac{2 x}{x ^{2} - 1} d x

ln x^{2} - 1 + C

Solution steps

\int \frac{2 x}{x ^{2} - 1} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x ^{2} - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

= 2 \cdot \frac{1}{2} ln x^{2} - 1

Simplify

2 \cdot \frac{1}{2} ln x^{2} - 1 : ln x^{2} - 1

= ln x^{2} - 1

Add a constant to the solution

= ln x^{2} - 1 + C

\int \frac{x ^{2}}{6 - x ^{3}} d x

\frac{d}{d x} (x^{4} - 9 x^{2} + 16)

\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}}

\int \frac{ln ( 4 x )}{x} d x

\frac{d}{d x} (x^{4} e^{5 x})

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