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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

u = arcsec (x)

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

Add a constant to the solution

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

4 cos^{2} (x)

1 9^{x}

\int 5 t^{2} e^{\frac{t ^{3}}{3}} d t

\int \frac{5 x ^{2}}{e ^{x}} d x

\frac{\partial}{\partial x} (e^{- 2 t} cos (2 x))

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