What is the integral of (48)/(x^4-10x^2+9) ?

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

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integral of (48)/(x^4-10x^2+9)

\int \frac{48}{x ^{4} - 10 x ^{2} + 9} d x

3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{48}{x ^{4} - 10 x ^{2} + 9} d x

Take the constant out:

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

= 48 \cdot \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 )}

= 48 \cdot \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

= 48 (\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 ∣

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

Simplify

48 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣) : 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣

= 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣ + C

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