What is the integral of ((x^4+6*x^2))/(((2+x^2)^2)) ?

The integral of ((x^4+6*x^2))/(((2+x^2)^2)) is x-1/(sqrt(2))sin(2arctan(x/(sqrt(2))))+C

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integral of ((x^4+6*x^2))/(((2+x^2)^2))

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

x - \frac{1}{2} sin (2 arctan (\frac{x}{2})) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4} + 6 x ^{2}}{( 2 + x ^{2} ) ^{2}} : 1 + \frac{2}{x ^{2} + 2} - \frac{8}{( x ^{2} + 2 ) ^{2}}

= \int 1 + \frac{2}{x ^{2} + 2} - \frac{8}{( x ^{2} + 2 ) ^{2}} d x

Apply the Sum Rule:

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

= \int 1 d x + \int \frac{2}{x ^{2} + 2} d x - \int \frac{8}{( x ^{2} + 2 ) ^{2}} d x

\int 1 d x = x

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

\int \frac{8}{( x ^{2} + 2 ) ^{2}} d x = 2 (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

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

Simplify

x + 2 arctan (\frac{x}{2}) - 2 (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : x - \frac{1}{2} sin (2 arctan (\frac{x}{2}))

= x - \frac{1}{2} sin (2 arctan (\frac{x}{2}))

Add a constant to the solution

= x - \frac{1}{2} sin (2 arctan (\frac{x}{2})) + C

\int \frac{x}{3} e^{2 x} d x

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\int_{0}^{1} x arctan (x) d x

What is the integral of ((x^4+6*x^2))/(((2+x^2)^2)) ?
The integral of ((x^4+6*x^2))/(((2+x^2)^2)) is x-1/(sqrt(2))sin(2arctan(x/(sqrt(2))))+C