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

The integral of (6x^2)/(x^4-5x^2+4) is ln|x+1|-ln|x-1|-2ln|x+2|+2ln|x-2|+C

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

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 5 x ^{2} + 4} : \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( x - 2 )}

= 6 \cdot \int \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( 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

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

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

\int \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} ln ∣ x + 2 ∣

\int \frac{1}{3 ( x - 2 )} d x = \frac{1}{3} ln ∣ x - 2 ∣

= 6 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 2 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) : ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

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What is the integral of (6x^2)/(x^4-5x^2+4) ?
The integral of (6x^2)/(x^4-5x^2+4) is ln|x+1|-ln|x-1|-2ln|x+2|+2ln|x-2|+C