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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} (x^{2} + 2 x + 3) e^{2 x} d x

y = ln (x^{4} (x))

f (x) = 3 x - x^{2}

\frac{d}{d x} (\frac{1}{ln ( ln ( ln ( x ) ) )})

f (x) = (2 x^{- 2} + 3)^{- 3}

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