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

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

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

\int \frac{6 x ^{3} - x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} d x

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

Solution steps

\int \frac{6 x ^{3} - x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} d x

Expand

\frac{6 x ^{3} - x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} : \frac{6 x ^{3}}{x ^{4} + x ^{2}} - \frac{x ^{2}}{x ^{4} + x ^{2}} + \frac{2 x}{x ^{4} + x ^{2}} - \frac{1}{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{6 x ^{3}}{x ^{4} + x ^{2}} d x - \int \frac{x ^{2}}{x ^{4} + x ^{2}} d x + \int \frac{2 x}{x ^{4} + x ^{2}} d x - \int \frac{1}{x ^{4} + x ^{2}} d x

\int \frac{6 x ^{3}}{x ^{4} + x ^{2}} d x = 3 ln x^{2} + 1

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

\int 3^{x} - 5 sin (2 x) d x

n = 0 \sum \infty (n + 1)^{2}

\frac{\partial}{\partial x} (\frac{x ^{2} + y}{y ^{2}})

y^{^{''}} - 4 y^{^{'}} + 8 y = 0

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