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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((2 x - x^{2}) (2 y - y^{2}))

\frac{\partial}{\partial x} (x^{yy - xx})

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

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

2 x y y^{^{'}} = 1 + y^{2}

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