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

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

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

\int \frac{9 ( x ^{4} - 1 )}{x ^{2} - 1} d x

9 (\frac{x ^{3}}{3} + x) + C

Solution steps

\int \frac{9 ( x ^{4} - 1 )}{x ^{2} - 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{4} - 1}{x ^{2} - 1} d x

Expand

\frac{x ^{4} - 1}{x ^{2} - 1} : \frac{x ^{4}}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int \frac{x ^{4}}{x ^{2} - 1} d x - \int \frac{1}{x ^{2} - 1} d x)

\int \frac{x ^{4}}{x ^{2} - 1} d x = \frac{x ^{3}}{3} + x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

\int \frac{1}{x ^{2} - 1} d x = - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

= 9 (\frac{x ^{3}}{3} + x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ - (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣))

Expand

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

= 9 (\frac{x ^{3}}{3} + x)

Add a constant to the solution

= 9 (\frac{x ^{3}}{3} + x) + C

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