What is the integral of 1/(sqrt(3x^2-2)) ?

The integral of 1/(sqrt(3x^2-2)) is 1/(sqrt(3))ln|sqrt(1/2 (3x^2-2))+sqrt(3/2)x|+C

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integral of 1/(sqrt(3x^2-2))

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

\frac{1}{3} ln \frac{1}{2} (3 x^{2} - 2) + \frac{3}{2} x + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

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

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= \frac{1}{3} ln tan (arcsec (\frac{3}{2} x)) + sec (arcsec (\frac{3}{2} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{2} x)) + sec (arcsec (\frac{3}{2} x)) : \frac{1}{3} ln \frac{1}{2} (3 x^{2} - 2) + \frac{3}{2} x

= \frac{1}{3} ln \frac{1}{2} (3 x^{2} - 2) + \frac{3}{2} x

Add a constant to the solution

= \frac{1}{3} ln \frac{1}{2} (3 x^{2} - 2) + \frac{3}{2} x + C

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

n = 1 \sum \infty \frac{n}{n + 1}

\int - t^{3} d t

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

\frac{d}{d x} (sin (x + 1) + 5 x)

What is the integral of 1/(sqrt(3x^2-2)) ?
The integral of 1/(sqrt(3x^2-2)) is 1/(sqrt(3))ln|sqrt(1/2 (3x^2-2))+sqrt(3/2)x|+C