What is the integral of 6/(xsqrt(x^2-1)) ?

The integral of 6/(xsqrt(x^2-1)) is 6arctan(sqrt(x^2-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 6/(xsqrt(x^2-1))

\int \frac{6}{x x ^{2} - 1} d x

6 arctan (x^{2} - 1) + C

Solution steps

\int \frac{6}{x x ^{2} - 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x x ^{2} - 1} d x

Apply u-substitution

= 6 \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 6 arctan (u)

Substitute back

u = x^{2} - 1

= 6 arctan (x^{2} - 1)

Add a constant to the solution

= 6 arctan (x^{2} - 1) + C

What is the integral of 6/(xsqrt(x^2-1)) ?
The integral of 6/(xsqrt(x^2-1)) is 6arctan(sqrt(x^2-1))+C