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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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