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

The integral of (4x-2)/(x^4+x^2) is 4(-1/2 ln|x^2+1|+ln(x))-2(-1/x-arctan(x))+C

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

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

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

Solution steps

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

Expand

\frac{4 x - 2}{x ^{4} + x ^{2}} : \frac{4 x}{x ^{4} + x ^{2}} - \frac{2}{x ^{4} + x ^{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 x}{x ^{4} + x ^{2}} d x - \int \frac{2}{x ^{4} + x ^{2}} d x

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

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

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

Add a constant to the solution

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

y = ln (x^{2} - 4)

x \to 2 lim (3 - x^{\frac{1}{x ^{2}}})

\frac{d}{d x} ((x) d x)

\frac{d}{d x} (cos (\frac{1 - e ^{6 x}}{1 + e ^{6 x}}))

x \to \infty lim (9.9)

What is the integral of (4x-2)/(x^4+x^2) ?
The integral of (4x-2)/(x^4+x^2) is 4(-1/2 ln|x^2+1|+ln(x))-2(-1/x-arctan(x))+C