What is the integral of 1/(x^2sqrt(4x^2-36)) ?

The integral of 1/(x^2sqrt(4x^2-36)) is (sqrt(x^2-9))/(18x)+C

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integral of 1/(x^2sqrt(4x^2-36))

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

\frac{x ^{2} - 9}{18 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{18 sec ( u )} d u

= \int \frac{1}{18 sec ( u )} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{18} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{18} sin (u)

Substitute back

u = arcsec (\frac{2}{6} x)

= \frac{1}{18} sin (arcsec (\frac{2}{6} x))

Simplify

\frac{1}{18} sin (arcsec (\frac{2}{6} x)) : \frac{x ^{2} - 9}{18 x}

= \frac{x ^{2} - 9}{18 x}

Add a constant to the solution

= \frac{x ^{2} - 9}{18 x} + C

\int (x + 5) 5 - x d x

f (x) = 8 x sin (x)

\frac{d}{d x} (ln (e^{2}))

\int_{0}^{3} (2 x + 1) d x

y^{^{''}} + 16 y = 16 sin (4 t)

What is the integral of 1/(x^2sqrt(4x^2-36)) ?
The integral of 1/(x^2sqrt(4x^2-36)) is (sqrt(x^2-9))/(18x)+C