What is the integral of 1/(x^2sqrt(4x^2-9)) ?

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

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integral of 1/(x^2sqrt(4x^2-9))

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

\frac{4 x ^{2} - 9}{9 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{2}{9 sec ( u )} d u

= \int \frac{2}{9 sec ( u )} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{2}{9} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{2}{9} sin (u)

Substitute back

u = arcsec (\frac{2}{3} x)

= \frac{2}{9} sin (arcsec (\frac{2}{3} x))

Simplify

\frac{2}{9} sin (arcsec (\frac{2}{3} x)) : \frac{4 x ^{2} - 9}{9 x}

= \frac{4 x ^{2} - 9}{9 x}

Add a constant to the solution

= \frac{4 x ^{2} - 9}{9 x} + C

\frac{\partial}{\partial y} (y - 2)

(1.01)^{x} + 0.3 3^{x}

4 \cdot (\frac{3}{2})^{x}

\frac{d}{d r} (\frac{2 k}{r} + 2 π r^{2})

y^{^{''}} - 4 y^{^{''}} + 2 y = 0

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