What is the integral of 1/(x(x^2-4)^{3/2)} ?

The integral of 1/(x(x^2-4)^{3/2)} is 1/8 (-arcsec(1/2 x)-2/((x^2-4)^{1/2)})+C

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integral of 1/(x(x^2-4)^{3/2)}

\int \frac{1}{x ( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

\frac{1}{8} (- arcsec (\frac{1}{2} x) - \frac{2}{( x ^{2} - 4 ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{1}{x ( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \frac{1}{8} \cdot \int - 1 + csc^{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} (- \int 1 d u + \int csc^{2} (u) d u)

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{8} (- u - cot (u))

Substitute back

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

= \frac{1}{8} (- arcsec (\frac{1}{2} x) - cot (arcsec (\frac{1}{2} x)))

Simplify

\frac{1}{8} (- arcsec (\frac{1}{2} x) - cot (arcsec (\frac{1}{2} x))) : \frac{1}{8} (- arcsec (\frac{1}{2} x) - \frac{2}{( x ^{2} - 4 ) ^{\frac{1}{2}}})

= \frac{1}{8} (- arcsec (\frac{1}{2} x) - \frac{2}{( x ^{2} - 4 ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{8} (- arcsec (\frac{1}{2} x) - \frac{2}{( x ^{2} - 4 ) ^{\frac{1}{2}}}) + C

\int_{0}^{20} 4 + 3 x d x

y^{^{''}} - 3 y^{^{'}} - 4 y = - 3, y (0) = 0, y^{^{'}} (0) = 0

\int tan^{2} (2 x se) c^{2} 2 x

y^{^{'}} + y - 2 cos (x) = 0

t^{10} e^{- 7 t}

What is the integral of 1/(x(x^2-4)^{3/2)} ?
The integral of 1/(x(x^2-4)^{3/2)} is 1/8 (-arcsec(1/2 x)-2/((x^2-4)^{1/2)})+C