What is the integral of 1/(sqrt(3y^2+4)) ?

The integral of 1/(sqrt(3y^2+4)) is 1/(sqrt(3))ln((|sqrt(3)y+sqrt(4+3y^2)|)/2)+C

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integral of 1/(sqrt(3y^2+4))

\int \frac{1}{3 y ^{2} + 4} d y

\frac{1}{3} ln \frac{3 y + 4 + 3 y ^{2}}{2} + C

Solution steps

\int \frac{1}{3 y ^{2} + 4} d y

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 = arctan (\frac{3}{2} y)

= \frac{1}{3} ln tan (arctan (\frac{3}{2} y)) + sec (arctan (\frac{3}{2} y))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2} y)) + sec (arctan (\frac{3}{2} y)) : \frac{1}{3} ln \frac{3 y + 4 + 3 y ^{2}}{2}

= \frac{1}{3} ln \frac{3 y + 4 + 3 y ^{2}}{2}

Add a constant to the solution

= \frac{1}{3} ln \frac{3 y + 4 + 3 y ^{2}}{2} + C

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What is the integral of 1/(sqrt(3y^2+4)) ?
The integral of 1/(sqrt(3y^2+4)) is 1/(sqrt(3))ln((|sqrt(3)y+sqrt(4+3y^2)|)/2)+C