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

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

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

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

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

Solution steps

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

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 (t)

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

Add a constant to the solution

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

\int_{0}^{1} x 1 + 8 x^{2} d x

\int (x^{2} - 4 x + 3) d x

\int sin^{4} (2 x) cos^{3} (2 x) d x

\frac{d y}{d x} = \frac{( x y + y ^{2} + x ^{2} )}{x ^{2}}

\frac{\partial}{\partial x} ((3 x + 4 y) e^{y})

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