What is the integral of sqrt(4y+1) ?

The integral of sqrt(4y+1) is 1/6 (4y+1)^{3/2}+C

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integral of sqrt(4y+1)

integral

4 y + 1

\frac{1}{6} (4 y + 1)^{\frac{3}{2}} + C

Solution steps

\int 4 y + 1 d y

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 y + 1

= \frac{1}{4} \cdot \frac{2}{3} (4 y + 1)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (4 y + 1)^{\frac{3}{2}} : \frac{1}{6} (4 y + 1)^{\frac{3}{2}}

= \frac{1}{6} (4 y + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (4 y + 1)^{\frac{3}{2}} + C

sec^{2} (y) tan (tan (y))

ln (4 - 2 x)

\frac{x}{( 2 + 2 x )}

\frac{1}{e ^{x} + 1}

\frac{( 2 x + 5 )}{( x + 2 )}

What is the integral of sqrt(4y+1) ?
The integral of sqrt(4y+1) is 1/6 (4y+1)^{3/2}+C