What is the integral of 1/(t^3sqrt(t^2-9)) ?

The integral of 1/(t^3sqrt(t^2-9)) is 1/54 (arcsec(1/3 t)+1/2 sin(2arcsec(1/3 t)))+C

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integral of 1/(t^3sqrt(t^2-9))

\int \frac{1}{t ^{3} t ^{2} - 9} d t

\frac{1}{54} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t))) + C

Solution steps

\int \frac{1}{t ^{3} t ^{2} - 9} d t

Apply Trigonometric Substitution

: \int \frac{1}{27 sec ^{2} ( u )} d u

= \int \frac{1}{27 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}{27} \cdot \int \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{27} \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}{27} \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}{27} \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}{27} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (\frac{1}{3} t)

= \frac{1}{27} \cdot \frac{1}{2} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t)))

Simplify

\frac{1}{27} \cdot \frac{1}{2} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t))) : \frac{1}{54} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t)))

= \frac{1}{54} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t)))

Add a constant to the solution

= \frac{1}{54} (arcsec (\frac{1}{3} t) + \frac{1}{2} sin (2 arcsec (\frac{1}{3} t))) + C

\int tan^{5} (2 x) sec^{2} (2 x) d x

\frac{4}{1 - 2 x ^{3}}

\int_{4}^{9} \frac{x}{9 + x ^{2}} d x

\int \frac{e ^{- x}}{x} d x

(11 y^{2} - x y) d x + x^{2} d y = 0

What is the integral of 1/(t^3sqrt(t^2-9)) ?
The integral of 1/(t^3sqrt(t^2-9)) is 1/54 (arcsec(1/3 t)+1/2 sin(2arcsec(1/3 t)))+C