What is the integral of 1/(sqrt(3x^2-5)) ?

The integral of 1/(sqrt(3x^2-5)) is 1/(sqrt(3))ln|sqrt(1/5 (3x^2-5))+sqrt(3/5)x|+C

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integral of 1/(sqrt(3x^2-5))

\int \frac{1}{3 x ^{2} - 5} d x

\frac{1}{3} ln \frac{1}{5} (3 x^{2} - 5) + \frac{3}{5} x + C

Solution steps

\int \frac{1}{3 x ^{2} - 5} 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}{5} x)

= \frac{1}{3} ln tan (arcsec (\frac{3}{5} x)) + sec (arcsec (\frac{3}{5} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{5} x)) + sec (arcsec (\frac{3}{5} x)) : \frac{1}{3} ln \frac{1}{5} (3 x^{2} - 5) + \frac{3}{5} x

= \frac{1}{3} ln \frac{1}{5} (3 x^{2} - 5) + \frac{3}{5} x

Add a constant to the solution

= \frac{1}{3} ln \frac{1}{5} (3 x^{2} - 5) + \frac{3}{5} x + C

\int \frac{8 x - 5}{3 x ^{2} - 5 x + 2} d x

\int \frac{3 a}{4 t} d t

\int \frac{1}{2 x ^{2} + 5 x + 3} d x

\int \frac{d}{1} (T) d x

\frac{1}{u ^{3}}

What is the integral of 1/(sqrt(3x^2-5)) ?
The integral of 1/(sqrt(3x^2-5)) is 1/(sqrt(3))ln|sqrt(1/5 (3x^2-5))+sqrt(3/5)x|+C