What is the integral of 1/(x^3sqrt(x^2-25)) ?

The integral of 1/(x^3sqrt(x^2-25)) is 1/250 (arcsec(1/5 x)+1/2 sin(2arcsec(1/5 x)))+C

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integral of 1/(x^3sqrt(x^2-25))

\int \frac{1}{x ^{3} x ^{2} - 25} d x

\frac{1}{250} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x))) + C

Solution steps

\int \frac{1}{x ^{3} x ^{2} - 25} d x

Apply Trigonometric Substitution

: \int \frac{1}{125 sec ^{2} ( u )} d u

= \int \frac{1}{125 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}{125} \cdot \int \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{125} \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}{125} \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}{125} \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}{125} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (\frac{1}{5} x)

= \frac{1}{125} \cdot \frac{1}{2} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x)))

Simplify

\frac{1}{125} \cdot \frac{1}{2} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x))) : \frac{1}{250} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x)))

= \frac{1}{250} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x)))

Add a constant to the solution

= \frac{1}{250} (arcsec (\frac{1}{5} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{5} x))) + C

\frac{t}{2 t - 3}

10 + \frac{20}{x + 1}, at x = 3

x \to \frac{π}{2} lim (cos (π + x))

n = 1 \sum \infty \frac{2}{5}

y^{^{'}} - y^{2} = 0, y (0) = 1

What is the integral of 1/(x^3sqrt(x^2-25)) ?
The integral of 1/(x^3sqrt(x^2-25)) is 1/250 (arcsec(1/5 x)+1/2 sin(2arcsec(1/5 x)))+C