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

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

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

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

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

Solution steps

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

Expand

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

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

Add a constant to the solution

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

y^{^{'}} - \frac{8}{x} y = \frac{y ^{5}}{x ^{11}}

\int_{1}^{2} (2 x + x^{2}) d x

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

x y^{2} \frac{d y}{d x} = y^{3} + x^{3}, y (1) = 5

\frac{\partial}{\partial x} (\frac{2 x}{x - y})

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