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

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

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integral of 1/(x(x^2-1)^{3/2)}

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \int cot^{2} (u) d u

Rewrite using trig identities

= \int - 1 + csc^{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

= - \int 1 d u + \int csc^{2} (u) d u

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= - u - cot (u)

Substitute back

u = arcsec (x)

= - arcsec (x) - cot (arcsec (x))

Simplify

- arcsec (x) - cot (arcsec (x)) : - arcsec (x) - \frac{1}{( x ^{2} - 1 ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

6 e^{x} (cos (x) - sin (x))

\int_{7}^{14} 196 - x^{2} d x

\int \frac{1}{3} e^{x + 3} + \frac{3}{x} - 5 d x

y^{^{'}} = \frac{y ^{2}}{x y + y ^{2}}

\frac{2 t}{2 + t}

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