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

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

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

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

ln ∣ x + 1 ∣ - arctan (x) + C

Solution steps

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

Expand

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

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

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

= ln ∣ x + 1 ∣ - arctan (x)

Add a constant to the solution

= ln ∣ x + 1 ∣ - arctan (x) + C

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