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

The integral of (x^3+3x-1)/(x^4-4x^2) is -3/4 ln|x|-1/(4x)+15/16 ln|x+2|+13/16 ln|x-2|+C

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

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

- \frac{3}{4} ln ∣ x ∣ - \frac{1}{4 x} + \frac{15}{16} ln ∣ x + 2 ∣ + \frac{13}{16} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} + 3 x - 1}{x ^{4} - 4 x ^{2}} : - \frac{3}{4 x} + \frac{1}{4 x ^{2}} + \frac{15}{16 ( x + 2 )} + \frac{13}{16 ( x - 2 )}

= \int - \frac{3}{4 x} + \frac{1}{4 x ^{2}} + \frac{15}{16 ( x + 2 )} + \frac{13}{16 ( x - 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 \frac{3}{4 x} d x + \int \frac{1}{4 x ^{2}} d x + \int \frac{15}{16 ( x + 2 )} d x + \int \frac{13}{16 ( x - 2 )} d x

\int \frac{3}{4 x} d x = \frac{3}{4} ln ∣ x ∣

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{15}{16 ( x + 2 )} d x = \frac{15}{16} ln ∣ x + 2 ∣

\int \frac{13}{16 ( x - 2 )} d x = \frac{13}{16} ln ∣ x - 2 ∣

= - \frac{3}{4} ln ∣ x ∣ - \frac{1}{4 x} + \frac{15}{16} ln ∣ x + 2 ∣ + \frac{13}{16} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{3}{4} ln ∣ x ∣ - \frac{1}{4 x} + \frac{15}{16} ln ∣ x + 2 ∣ + \frac{13}{16} ln ∣ x - 2 ∣ + C

x \to 0 - lim (\frac{e ^{3 x}}{2 x})

\frac{d y}{d x} = x y^{4} - y

x \to 1 - f lim (x)

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

y = - x^{3}, x = - 2, x = 2, y = 0

What is the integral of (x^3+3x-1)/(x^4-4x^2) ?
The integral of (x^3+3x-1)/(x^4-4x^2) is -3/4 ln|x|-1/(4x)+15/16 ln|x+2|+13/16 ln|x-2|+C