What is the integral of 1/(t^3sqrt(t^2-4)) ?

The integral of 1/(t^3sqrt(t^2-4)) is 1/16 (arcsec(1/2 t)+1/2 sin(2arcsec(1/2 t)))+C

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integral of 1/(t^3sqrt(t^2-4))

\int \frac{1}{t ^{3} t ^{2} - 4} d t

\frac{1}{16} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t))) + C

Solution steps

\int \frac{1}{t ^{3} t ^{2} - 4} d t

Apply Trigonometric Substitution

: \int \frac{1}{8 sec ^{2} ( u )} d u

= \int \frac{1}{8 sec ^{2} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{8} \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

= \frac{1}{8} \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

= \frac{1}{8} \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)

= \frac{1}{8} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (\frac{1}{2} t)

= \frac{1}{8} \cdot \frac{1}{2} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t)))

Simplify

\frac{1}{8} \cdot \frac{1}{2} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t))) : \frac{1}{16} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t)))

= \frac{1}{16} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t)))

Add a constant to the solution

= \frac{1}{16} (arcsec (\frac{1}{2} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{2} t))) + C

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

\int_{1}^{4} ln (x) d x

f (x) = x^{2} + 6 x - 5, at x = 2

\int 12 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

\frac{x}{x + 3}

What is the integral of 1/(t^3sqrt(t^2-4)) ?
The integral of 1/(t^3sqrt(t^2-4)) is 1/16 (arcsec(1/2 t)+1/2 sin(2arcsec(1/2 t)))+C