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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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