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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{18 u} d u

= \int \frac{1}{18 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{18} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 9 y^{2} + 4

= \frac{1}{18} \cdot 2 (9 y^{2} + 4)^{\frac{1}{2}}

Simplify

\frac{1}{18} \cdot 2 (9 y^{2} + 4)^{\frac{1}{2}} : \frac{1}{9} 9 y^{2} + 4

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

Add a constant to the solution

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

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