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

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

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

\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 ∣ + C

Solution steps

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

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

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

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

Add a constant to the solution

= \frac{x ^{3}}{3} + x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

\int x cos (x n) d x

\int x (6 - x) d x

\frac{s}{( s ^{2} - k ^{2} )}

\frac{\partial}{\partial x} ((x^{3}) x^{2} + 9)

x \to 1 - lim (\frac{x ^{2} - 1}{x - 1})

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