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

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

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

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

arcsec (x) + \frac{1}{2} sin (2 arcsec (x)) + C

Solution steps

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

Apply Trigonometric Substitution

: \int cos^{2} (u) d u

= 2 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int \frac{1 + cos ( 2 u )}{2} 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 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 2 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (x)

= 2 \cdot \frac{1}{2} (arcsec (x) + \frac{1}{2} sin (2 arcsec (x)))

Simplify

2 \cdot \frac{1}{2} (arcsec (x) + \frac{1}{2} sin (2 arcsec (x))) : arcsec (x) + \frac{1}{2} sin (2 arcsec (x))

= arcsec (x) + \frac{1}{2} sin (2 arcsec (x))

Add a constant to the solution

= arcsec (x) + \frac{1}{2} sin (2 arcsec (x)) + C

\int (2 x - y) d x

\frac{\partial}{\partial y} (y^{2} x)

\frac{d}{d x} (0.00037 x^{2})

ln (1 + 5 x)

\int_{0}^{1} x 1 - x^{4} d x

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