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

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

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

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

- ln ∣ x ∣ - \frac{2}{x} - ln x^{2} + 2 + C

Solution steps

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

Take the partial fraction of

\frac{3 x ^{3} - 2 x ^{2} + 2 x - 4}{- x ^{4} - 2 x ^{2}} : - \frac{x - 2}{x ^{2}} - \frac{2 x}{x ^{2} + 2}

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

\int \frac{x - 2}{x ^{2}} d x = ln ∣ x ∣ + \frac{2}{x}

\int \frac{2 x}{x ^{2} + 2} d x = ln x^{2} + 2

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

Simplify

= - ln ∣ x ∣ - \frac{2}{x} - ln x^{2} + 2

Add a constant to the solution

= - ln ∣ x ∣ - \frac{2}{x} - ln x^{2} + 2 + C

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

\int_{1}^{2} (\frac{3}{x ^{2}} - 1) d x

\frac{d y}{d x} = \frac{2 x + y + 1}{4 x + 2 y - 5}

x \to 0 lim (5 + 4 x - x^{3})

x \to - \infty lim (4)

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