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

The integral of (3x^2-x+2)/((x^2+1)^2(x+1)) is -3/4 ln|x^2+1|+arctan(x)-x/(2(1+x^2))+3/2 ln|x+1|+C

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (3x^2-x+2)/((x^2+1)^2(x+1)) ?
The integral of (3x^2-x+2)/((x^2+1)^2(x+1)) is -3/4 ln|x^2+1|+arctan(x)-x/(2(1+x^2))+3/2 ln|x+1|+C