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

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

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

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

3 ln ∣ x + 1 ∣ - arctan (x) + C

Solution steps

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

Expand

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

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

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

Simplify

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

= 3 ln ∣ x + 1 ∣ - arctan (x)

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - arctan (x) + C

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