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

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

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integral of sqrt(1-1/(x^2)) 1/(x^3)

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

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

Solution steps

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

Simplify

1 - \frac{1}{x ^{2}} \frac{1}{x ^{3}} : \frac{x ^{2} - 1}{x ^{3} x ^{2}}

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

x^{2} = (x),

assuming

x \geq 0

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

x^{3} x = x^{4}

= \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

= - \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

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