What is the integral of (sqrt(x^2-1))/(x^4) ?

The integral of (sqrt(x^2-1))/(x^4) is -(sqrt(x^2-1))/(3x^3)+(sqrt(x^2-1))/(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (sqrt(x^2-1))/(x^4)

\int \frac{x ^{2} - 1}{x ^{4}} d x

- \frac{x ^{2} - 1}{3 x ^{3}} + \frac{x ^{2} - 1}{3 x} + C

Solution steps

\int \frac{x ^{2} - 1}{x ^{4}} d x

Apply Integration By Parts

= - \frac{x ^{2} - 1}{3 x ^{3}} - \int - \frac{1}{3 x ^{2} x ^{2} - 1} d x

\int - \frac{1}{3 x ^{2} x ^{2} - 1} d x = - \frac{x ^{2} - 1}{3 x}

= - \frac{x ^{2} - 1}{3 x ^{3}} - (- \frac{x ^{2} - 1}{3 x})

Simplify

= - \frac{x ^{2} - 1}{3 x ^{3}} + \frac{x ^{2} - 1}{3 x}

Add a constant to the solution

= - \frac{x ^{2} - 1}{3 x ^{3}} + \frac{x ^{2} - 1}{3 x} + C

What is the integral of (sqrt(x^2-1))/(x^4) ?
The integral of (sqrt(x^2-1))/(x^4) is -(sqrt(x^2-1))/(3x^3)+(sqrt(x^2-1))/(3x)+C