What is the integral of (2x^2-5)/(x^4-5x^2+6) ?

The integral of (2x^2-5)/(x^4-5x^2+6) is

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integral of (2x^2-5)/(x^4-5x^2+6)

\int \frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} d x

2 ln x + 2 - 2 ln x - 2 - 3 ln x + 3 + 3 ln x - 3 - 5 (\frac{1}{2 2} ln x + 2 - \frac{1}{2 2} ln x - 2 - \frac{1}{2 3} ln x + 3 + \frac{1}{2 3} ln x - 3) + C

Solution steps

\int \frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} d x

Expand

\frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} : \frac{2 x ^{2}}{x ^{4} - 5 x ^{2} + 6} - \frac{5}{x ^{4} - 5 x ^{2} + 6}

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{2 x ^{2}}{x ^{4} - 5 x ^{2} + 6} d x - \int \frac{5}{x ^{4} - 5 x ^{2} + 6} d x

\int \frac{2 x ^{2}}{x ^{4} - 5 x ^{2} + 6} d x = 2 ln x + 2 - 2 ln x - 2 - 3 ln x + 3 + 3 ln x - 3

\int \frac{5}{x ^{4} - 5 x ^{2} + 6} d x = 5 (\frac{1}{2 2} ln x + 2 - \frac{1}{2 2} ln x - 2 - \frac{1}{2 3} ln x + 3 + \frac{1}{2 3} ln x - 3)

= 2 ln x + 2 - 2 ln x - 2 - 3 ln x + 3 + 3 ln x - 3 - 5 (\frac{1}{2 2} ln x + 2 - \frac{1}{2 2} ln x - 2 - \frac{1}{2 3} ln x + 3 + \frac{1}{2 3} ln x - 3)

Add a constant to the solution

= 2 ln x + 2 - 2 ln x - 2 - 3 ln x + 3 + 3 ln x - 3 - 5 (\frac{1}{2 2} ln x + 2 - \frac{1}{2 2} ln x - 2 - \frac{1}{2 3} ln x + 3 + \frac{1}{2 3} ln x - 3) + C

\int (2 x^{2} + 2 x - 3)^{10} (2 x + 1) d x

\frac{\partial}{\partial x} (5 x^{7} cos (y^{8}))

x \to 0 - lim (\frac{7}{x} - \frac{7}{∣ x ∣})

\int_{0}^{1} \frac{1}{1 - x ^{2}} d x

x^{4} - 17 x^{2} + 16

What is the integral of (2x^2-5)/(x^4-5x^2+6) ?
The integral of (2x^2-5)/(x^4-5x^2+6) is