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

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

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

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

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

Solution steps

\int \frac{2 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

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

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 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{4} \frac{1}{x ^{2} + 16} d x

\frac{x}{4 + x}

\frac{\partial}{\partial x} (x y + x e^{\frac{y}{x}})

\int 8 t e^{t} d t

4 \frac{d y}{d x} + 20 y = 5

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