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

The integral of (10)/((x-1)(x^2-9)) is 10(-1/8 ln|x-1|+1/24 ln|x+3|+1/12 ln|x-3|)+C

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

\int \frac{10}{( x - 1 ) ( x ^{2} - 9 )} d x

10 (- \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{10}{( x - 1 ) ( x ^{2} - 9 )} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} - 9 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} - 9 )} : - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( x - 3 )}

= 10 \cdot \int - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( 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

= 10 (- \int \frac{1}{8 ( x - 1 )} d x + \int \frac{1}{24 ( x + 3 )} d x + \int \frac{1}{12 ( x - 3 )} d x)

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

\int \frac{1}{24 ( x + 3 )} d x = \frac{1}{24} ln ∣ x + 3 ∣

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

= 10 (- \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

Add a constant to the solution

= 10 (- \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣) + C

x \to 0 - lim (8 - x^{2})

f (x) = 1 + cos (x), at x = \frac{π}{2}

\frac{d x}{d t} + t^{5} x^{7} + \frac{x}{t} = 0

f (x) = x cos (x)

\int \frac{1}{( x + 1 ) x} d x

What is the integral of (10)/((x-1)(x^2-9)) ?
The integral of (10)/((x-1)(x^2-9)) is 10(-1/8 ln|x-1|+1/24 ln|x+3|+1/12 ln|x-3|)+C