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

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

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

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

x + ln x^{3} + 1 + C

Solution steps

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

Expand

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

\int \frac{x ^{3}}{x ^{3} + 1} d x = - \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) + x

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

\int \frac{1}{x ^{3} + 1} d x = \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))

= - \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) + x + ln x^{3} + 1 + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))

Simplify

- \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) + x + ln x^{3} + 1 + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) : x + ln x^{3} + 1

= x + ln x^{3} + 1

Add a constant to the solution

= x + ln x^{3} + 1 + C

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