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

The integral of 2/(x^2(x^2-4x-5)) is 2(4/25 ln|x|+1/(5x)-1/6 ln|x+1|+1/150 ln|x-5|)+C

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

\int \frac{2}{x ^{2} ( x ^{2} - 4 x - 5 )} d x

2 (\frac{4}{25} ln ∣ x ∣ + \frac{1}{5 x} - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{150} ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{2}{x ^{2} ( x ^{2} - 4 x - 5 )} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} ( x ^{2} - 4 x - 5 )} d x

Take the partial fraction of

\frac{1}{x ^{2} ( x ^{2} - 4 x - 5 )} : \frac{4}{25 x} - \frac{1}{5 x ^{2}} - \frac{1}{6 ( x + 1 )} + \frac{1}{150 ( x - 5 )}

= 2 \cdot \int \frac{4}{25 x} - \frac{1}{5 x ^{2}} - \frac{1}{6 ( x + 1 )} + \frac{1}{150 ( x - 5 )} 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

= 2 (\int \frac{4}{25 x} d x - \int \frac{1}{5 x ^{2}} d x - \int \frac{1}{6 ( x + 1 )} d x + \int \frac{1}{150 ( x - 5 )} d x)

\int \frac{4}{25 x} d x = \frac{4}{25} ln ∣ x ∣

\int \frac{1}{5 x ^{2}} d x = - \frac{1}{5 x}

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

\int \frac{1}{150 ( x - 5 )} d x = \frac{1}{150} ln ∣ x - 5 ∣

= 2 (\frac{4}{25} ln ∣ x ∣ - (- \frac{1}{5 x}) - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{150} ln ∣ x - 5 ∣)

Simplify

= 2 (\frac{4}{25} ln ∣ x ∣ + \frac{1}{5 x} - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{150} ln ∣ x - 5 ∣)

Add a constant to the solution

= 2 (\frac{4}{25} ln ∣ x ∣ + \frac{1}{5 x} - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{150} ln ∣ x - 5 ∣) + C

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What is the integral of 2/(x^2(x^2-4x-5)) ?
The integral of 2/(x^2(x^2-4x-5)) is 2(4/25 ln|x|+1/(5x)-1/6 ln|x+1|+1/150 ln|x-5|)+C