What is the integral of (6x)/((4-3x^2)^4) ?

The integral of (6x)/((4-3x^2)^4) is -1/(3(-4+3x^2)^3)+C

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integral of (6x)/((4-3x^2)^4)

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

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

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 4 - 3 x ^{2} ) ^{4}} d x

Factor

4 - 3 x^{2} : - (- 4 + 3 x^{2})

= 6 \cdot \int \frac{x}{( - ( - 4 + 3 x ^{2} ) ) ^{4}} d x

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

= 6 \cdot \int \frac{x}{( - 4 + 3 x ^{2} ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{6 u ^{4}} d u

= 6 \cdot \int \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = - 4 + 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (\frac{x ^{2} + 4}{x ^{2} + 1}))

x^{3} - 6 x^{2} - 15 x + 4

x \to 5 lim (x^{- 2})

y^{^{'}} = \frac{9 x}{y + x ^{2} y}

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

What is the integral of (6x)/((4-3x^2)^4) ?
The integral of (6x)/((4-3x^2)^4) is -1/(3(-4+3x^2)^3)+C