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

The integral of y/(sqrt(3y+1)) is 2/9 (1/3 (3y+1)^{3/2}-sqrt(3y+1))+C

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integral of y/(sqrt(3y+1))

\int \frac{y}{3 y + 1} d y

\frac{2}{9} (\frac{1}{3} (3 y + 1)^{\frac{3}{2}} - 3 y + 1) + C

Solution steps

\int \frac{y}{3 y + 1} d y

Apply u-substitution

: \int \frac{2 ( u ^{2} - 1 )}{9} d u

= \int \frac{2 ( u ^{2} - 1 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{9} (\int u^{2} d u - \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{2}{9} (\frac{u ^{3}}{3} - u)

Substitute back

u = 3 y + 1

= \frac{2}{9} (\frac{( 3 y + 1 ) ^{3}}{3} - 3 y + 1)

Simplify

\frac{2}{9} (\frac{( 3 y + 1 ) ^{3}}{3} - 3 y + 1) : \frac{2}{9} (\frac{1}{3} (3 y + 1)^{\frac{3}{2}} - 3 y + 1)

= \frac{2}{9} (\frac{1}{3} (3 y + 1)^{\frac{3}{2}} - 3 y + 1)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (3 y + 1)^{\frac{3}{2}} - 3 y + 1) + C

\frac{d y}{d x} = 3 cos (x) y, y (\frac{π}{2}) = - 2

\frac{d}{d x} ((x^{2}) (\frac{2}{x ^{2}} + \frac{5}{x ^{3}}))

x \to - \infty lim (0. 1^{x})

\frac{d}{d x} (x^{\frac{4}{5}} (x - 2)^{2})

\frac{d}{d x} (\frac{( x ^{2} )}{1 + x})

What is the integral of y/(sqrt(3y+1)) ?
The integral of y/(sqrt(3y+1)) is 2/9 (1/3 (3y+1)^{3/2}-sqrt(3y+1))+C