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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

k = 1 \sum \infty k^{3} \cdot e^{- k^{2}}

\frac{\partial}{\partial x} (4 x^{6} + x y^{3} + 2 y^{4})

\frac{\partial}{\partial y} (y ln (x^{2} y))

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

\int 1 - sin^{2} (θ) d θ

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