What is the integral of 1/(sqrt(9y^2+4)) ?

The integral of 1/(sqrt(9y^2+4)) is 1/3 ln((|3y+sqrt(4+9y^2)|)/2)+C

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integral of 1/(sqrt(9y^2+4))

\int \frac{1}{9 y ^{2} + 4} d y

\frac{1}{3} ln \frac{3 y + 4 + 9 y ^{2}}{2} + C

Solution steps

\int \frac{1}{9 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 + 9 y ^{2}}{2}

= \frac{1}{3} ln \frac{3 y + 4 + 9 y ^{2}}{2}

Add a constant to the solution

= \frac{1}{3} ln \frac{3 y + 4 + 9 y ^{2}}{2} + C

\int (1 - x) d x

e^{- 4 t} \times sin (2 t)

\frac{10}{3} x^{3} - 5 x^{2} - 11 x + 6

\int \frac{x}{100 x ^{2} - 1} d x

\frac{d y}{d x}, x = y^{18}

What is the integral of 1/(sqrt(9y^2+4)) ?
The integral of 1/(sqrt(9y^2+4)) is 1/3 ln((|3y+sqrt(4+9y^2)|)/2)+C