What is the integral of 1/(3xsqrt(9x^2-12)) ?

The integral of 1/(3xsqrt(9x^2-12)) is 1/(6sqrt(3))arcsec((sqrt(3)x)/2)+C

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integral of 1/(3xsqrt(9x^2-12))

\int \frac{1}{3 x 9 x ^{2} - 12} d x

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

Solution steps

\int \frac{1}{3 x 9 x ^{2} - 12} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{x 9 x ^{2} - 12} d x

Apply Trigonometric Substitution

: \int \frac{1}{2 3} d u

= \frac{1}{3} \cdot \int \frac{1}{2 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 \frac{1}{2 3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot \frac{1}{2 3} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

e^{- 1}

\int_{0}^{2} e^{- x} + e d x

\int t e^{- 6 t} d t

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x})

n = 1 \sum \infty 4 (0.8)^{n - 1}

What is the integral of 1/(3xsqrt(9x^2-12)) ?
The integral of 1/(3xsqrt(9x^2-12)) is 1/(6sqrt(3))arcsec((sqrt(3)x)/2)+C