What is the integral of 1/(x^4-10x^2+9) ?

The integral of 1/(x^4-10x^2+9) is 1/16 ln|x+1|-1/16 ln|x-1|-1/48 ln|x+3|+1/48 ln|x-3|+C

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integral of 1/(x^4-10x^2+9)

\int \frac{1}{x ^{4} - 10 x ^{2} + 9} d x

\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{x ^{4} - 10 x ^{2} + 9} d x

Take the partial fraction of

\frac{1}{x ^{4} - 10 x ^{2} + 9} : \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} - \frac{1}{48 ( x + 3 )} + \frac{1}{48 ( x - 3 )}

= \int \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} - \frac{1}{48 ( x + 3 )} + \frac{1}{48 ( x - 3 )} 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{1}{16 ( x + 1 )} d x - \int \frac{1}{16 ( x - 1 )} d x - \int \frac{1}{48 ( x + 3 )} d x + \int \frac{1}{48 ( x - 3 )} d x

\int \frac{1}{16 ( x + 1 )} d x = \frac{1}{16} ln ∣ x + 1 ∣

\int \frac{1}{16 ( x - 1 )} d x = \frac{1}{16} ln ∣ x - 1 ∣

\int \frac{1}{48 ( x + 3 )} d x = \frac{1}{48} ln ∣ x + 3 ∣

\int \frac{1}{48 ( x - 3 )} d x = \frac{1}{48} ln ∣ x - 3 ∣

= \frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣ + C

x \to 0 lim ((1 - 3 x)^{\frac{4}{x}})

x \to \infty lim (x^{2} + 6 x)

\int_{- 1}^{1} ∣ 2 x - 1 ∣ d x

\frac{d}{d x} (2 tan (cos (3 x)))

\int sin (29 x) d x

What is the integral of 1/(x^4-10x^2+9) ?
The integral of 1/(x^4-10x^2+9) is 1/16 ln|x+1|-1/16 ln|x-1|-1/48 ln|x+3|+1/48 ln|x-3|+C