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

The integral of (3y)/((2+3y^2)^3) is -1/(4(3y^2+2)^2)+C

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

\int \frac{3 y}{( 2 + 3 y ^{2} ) ^{3}} d y

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

Solution steps

\int \frac{3 y}{( 2 + 3 y ^{2} ) ^{3}} 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 + 3 y ^{2} ) ^{3}} d y

Apply u-substitution

: \int \frac{1}{6 u ^{3}} d u

= 3 \cdot \int \frac{1}{6 u ^{3}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{6} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 + 3 y^{2}

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

Simplify

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

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

Add a constant to the solution

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

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

\int 1 + sin (x) d x

f (x) = 2^{x + 1}

f (x) = 5 + 4 e^{3 x}

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = - 3, y^{^{'}} (0) = 28

What is the integral of (3y)/((2+3y^2)^3) ?
The integral of (3y)/((2+3y^2)^3) is -1/(4(3y^2+2)^2)+C