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

The integral of (2y)/((2+3y^2)^3) is -1/(6(2+3y^2)^2)+C

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

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

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

Solution steps

\int \frac{2 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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 + 3 y^{2}

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

Simplify

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

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

Add a constant to the solution

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

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x \to - 1 lim (- x^{3} + x^{2} - 2)

\int 6 x^{5} d x

\frac{d}{d x} (8 e^{- 4 x})

f (x) = 8 x^{4} tan (x)

What is the integral of (2y)/((2+3y^2)^3) ?
The integral of (2y)/((2+3y^2)^3) is -1/(6(2+3y^2)^2)+C