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

The integral of (6x+4)/(x^2(x^2+2)) is

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

(x - 2 y) d y + (y + 4 x) d x = 0

3 x^{4} 25 - x^{2}

\int 4 sin (t) cos (t) d t

x \to 3 lim (\frac{x ^{3} - 27}{x - 3})

\frac{d y}{d x} = \frac{3 sec ( y )}{( x - 4 ) ^{2}}

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