What is the integral of (3y)/(sqrt(2y^2+5)) ?

The integral of (3y)/(sqrt(2y^2+5)) is 3/2 sqrt(2y^2+5)+C

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integral of (3y)/(sqrt(2y^2+5))

\int \frac{3 y}{2 y ^{2} + 5} d y

\frac{3}{2} 2 y^{2} + 5 + C

Solution steps

\int \frac{3 y}{2 y ^{2} + 5} d y

Take the constant out:

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

= 3 \cdot \int \frac{y}{2 y ^{2} + 5} d y

Apply u-substitution

: \int \frac{1}{4 u} d u

= 3 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

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

Substitute back

u = 2 y^{2} + 5

= 3 \cdot \frac{1}{4} \cdot 2 (2 y^{2} + 5)^{\frac{1}{2}}

Simplify

3 \cdot \frac{1}{4} \cdot 2 (2 y^{2} + 5)^{\frac{1}{2}} : \frac{3}{2} 2 y^{2} + 5

= \frac{3}{2} 2 y^{2} + 5

Add a constant to the solution

= \frac{3}{2} 2 y^{2} + 5 + C

arctan (8 x)

\frac{d}{d x} (2 x^{5} - 6 x^{4} + 4 x^{3} + 4 x^{2} - 6 x + 2)

\int (x - 2) d x

\frac{d}{d x} (\frac{3 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}})

y^{^{''}} + 2 y^{^{'}} + y = - 2 e^{x}

What is the integral of (3y)/(sqrt(2y^2+5)) ?
The integral of (3y)/(sqrt(2y^2+5)) is 3/2 sqrt(2y^2+5)+C