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

The integral of ((4-3x^2))/((3x^2+4)^2) is x/(2(3x^2+4))+1/(8sqrt(3))sin(2arctan((sqrt(3))/2 x))+C

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

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

\frac{x}{2 ( 3 x ^{2} + 4 )} + \frac{1}{8 3} sin (2 arctan (\frac{3}{2} x)) + C

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{4}{( 3 x ^{2} + 4 ) ^{2}} d x = \frac{1}{4 3} (arctan (\frac{3}{2} x) + \frac{1}{2} sin (2 arctan (\frac{3}{2} x)))

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

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

Simplify

\frac{1}{4 3} (arctan (\frac{3}{2} x) + \frac{1}{2} sin (2 arctan (\frac{3}{2} x))) - (\frac{1}{4 3} arctan (\frac{3}{2} x) - \frac{x}{2 ( 4 + 3 x ^{2} )}) : \frac{x}{2 ( 3 x ^{2} + 4 )} + \frac{1}{8 3} sin (2 arctan (\frac{3}{2} x))

= \frac{x}{2 ( 3 x ^{2} + 4 )} + \frac{1}{8 3} sin (2 arctan (\frac{3}{2} x))

Add a constant to the solution

= \frac{x}{2 ( 3 x ^{2} + 4 )} + \frac{1}{8 3} sin (2 arctan (\frac{3}{2} x)) + C

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

\frac{\partial}{\partial z} (x \cdot sin (y) \cdot sin (z))

\frac{\partial}{\partial y} (\frac{y}{x} + 6 x)

\int \frac{1}{8 x ^{4}} d x

3^{x^{2} - 1}, at x = 2

What is the integral of ((4-3x^2))/((3x^2+4)^2) ?
The integral of ((4-3x^2))/((3x^2+4)^2) is x/(2(3x^2+4))+1/(8sqrt(3))sin(2arctan((sqrt(3))/2 x))+C