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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

x \to 5 lim (2 x - 7)

\int \frac{1}{π} x cos (n x) d x

x \to - 4 lim (\frac{4}{x + 4})

\frac{\partial}{\partial x} (3 x^{2} + x y^{3} - 2 y^{2} + 3)

y = tan (x), at x = - \frac{7 π}{4}

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