What is the integral of 1/(x(x^2-36)^{3/2)} ?

The integral of 1/(x(x^2-36)^{3/2)} is 1/216 (-arcsec(1/6 x)-6/((x^2-36)^{1/2)})+C

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integral of 1/(x(x^2-36)^{3/2)}

\int \frac{1}{x ( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

\frac{1}{216} (- arcsec (\frac{1}{6} x) - \frac{6}{( x ^{2} - 36 ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{1}{x ( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{216 tan ^{2} ( u )} d u

= \int \frac{1}{216 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}{216} \cdot \int \frac{1}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{216} \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}{216} (- \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}{216} (- u - cot (u))

Substitute back

u = arcsec (\frac{1}{6} x)

= \frac{1}{216} (- arcsec (\frac{1}{6} x) - cot (arcsec (\frac{1}{6} x)))

Simplify

\frac{1}{216} (- arcsec (\frac{1}{6} x) - cot (arcsec (\frac{1}{6} x))) : \frac{1}{216} (- arcsec (\frac{1}{6} x) - \frac{6}{( x ^{2} - 36 ) ^{\frac{1}{2}}})

= \frac{1}{216} (- arcsec (\frac{1}{6} x) - \frac{6}{( x ^{2} - 36 ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{216} (- arcsec (\frac{1}{6} x) - \frac{6}{( x ^{2} - 36 ) ^{\frac{1}{2}}}) + C

3 cos^{3} (t)

(ln (y))^{6} \frac{d y}{d x} = x^{6} y, y (1) = e^{2}

y = e^{x} (3 - 5 x + 3 x^{2})

n = 0 \sum \infty n^{2} + 1

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

What is the integral of 1/(x(x^2-36)^{3/2)} ?
The integral of 1/(x(x^2-36)^{3/2)} is 1/216 (-arcsec(1/6 x)-6/((x^2-36)^{1/2)})+C