What is the integral of (-9)/(sqrt(x^4-36x^2)) ?

The integral of (-9)/(sqrt(x^4-36x^2)) is -3/2 arcsec(1/6 x)+C

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integral of (-9)/(sqrt(x^4-36x^2))

\int \frac{- 9}{x ^{4} - 36 x ^{2}} d x

- \frac{3}{2} arcsec (\frac{1}{6} x) + C

Solution steps

\int \frac{- 9}{x ^{4} - 36 x ^{2}} d x

Take the constant out:

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

= - 9 \cdot \int \frac{1}{x ^{4} - 36 x ^{2}} d x

x^{4} - 36 x^{2} = x^{2} x^{2} - 36

= - 9 \cdot \int \frac{1}{x ^{2} x ^{2} - 36} d x

x^{2} = (x),

assuming

x \geq 0

= - 9 \cdot \int \frac{1}{x x ^{2} - 36} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= - 9 \cdot \int \frac{1}{6} d u

Take the constant out:

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

= - 9 \cdot \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - 9 \cdot \frac{1}{6} \cdot 1 \cdot u

Substitute back

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

= - 9 \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} x)

Simplify

- 9 \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} x) : - \frac{3}{2} arcsec (\frac{1}{6} x)

= - \frac{3}{2} arcsec (\frac{1}{6} x)

Add a constant to the solution

= - \frac{3}{2} arcsec (\frac{1}{6} x) + C

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What is the integral of (-9)/(sqrt(x^4-36x^2)) ?
The integral of (-9)/(sqrt(x^4-36x^2)) is -3/2 arcsec(1/6 x)+C