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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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