What is the integral of (x^3-6x^2+4x-10)/(x^2-1) ?

The integral of (x^3-6x^2+4x-10)/(x^2-1) is 1/2 x^2-6x+5/2 ln|x^2-1|+8ln|x+1|-8ln|x-1|-1/2+C

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integral of (x^3-6x^2+4x-10)/(x^2-1)

\int \frac{x ^{3} - 6 x ^{2} + 4 x - 10}{x ^{2} - 1} d x

\frac{1}{2} x^{2} - 6 x + \frac{5}{2} ln x^{2} - 1 + 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ - \frac{1}{2} + C

Solution steps

\int \frac{x ^{3} - 6 x ^{2} + 4 x - 10}{x ^{2} - 1} d x

Expand

\frac{x ^{3} - 6 x ^{2} + 4 x - 10}{x ^{2} - 1} : \frac{x ^{3}}{x ^{2} - 1} - \frac{6 x ^{2}}{x ^{2} - 1} + \frac{4 x}{x ^{2} - 1} - \frac{10}{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

= \int \frac{x ^{3}}{x ^{2} - 1} d x - \int \frac{6 x ^{2}}{x ^{2} - 1} d x + \int \frac{4 x}{x ^{2} - 1} d x - \int \frac{10}{x ^{2} - 1} d x

\int \frac{x ^{3}}{x ^{2} - 1} d x = \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1)

\int \frac{6 x ^{2}}{x ^{2} - 1} d x = - 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)

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

\int \frac{10}{x ^{2} - 1} d x = - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) + 2 ln x^{2} - 1 - (- 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣))

Simplify

\frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) + 2 ln x^{2} - 1 - (- 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)) : \frac{1}{2} x^{2} - 6 x + \frac{5}{2} ln x^{2} - 1 + 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ - \frac{1}{2}

= \frac{1}{2} x^{2} - 6 x + \frac{5}{2} ln x^{2} - 1 + 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ - \frac{1}{2}

Add a constant to the solution

= \frac{1}{2} x^{2} - 6 x + \frac{5}{2} ln x^{2} - 1 + 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ - \frac{1}{2} + C

\int cos (4 x) cos (7 x) d x

f (x) = - \frac{6}{x ^{4}}

x \to \frac{π}{4} lim (cos (x))

n \to \infty lim (n 10 n)

3 y^{^{'}} + y = 0

What is the integral of (x^3-6x^2+4x-10)/(x^2-1) ?
The integral of (x^3-6x^2+4x-10)/(x^2-1) is 1/2 x^2-6x+5/2 ln|x^2-1|+8ln|x+1|-8ln|x-1|-1/2+C