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

The integral of (3x^2-x+1)/((x+1)(x^2+2x+2)) is 5ln|x+1|-ln|x^2+2x+2|-7arctan(x+1)+C

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

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

5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C

Solution steps

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

Take the partial fraction of

\frac{3 x ^{2} - x + 1}{( x + 1 ) ( x ^{2} + 2 x + 2 )} : \frac{5}{x + 1} + \frac{- 2 x - 9}{x ^{2} + 2 x + 2}

= \int \frac{5}{x + 1} + \frac{- 2 x - 9}{x ^{2} + 2 x + 2} 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{5}{x + 1} d x + \int \frac{- 2 x - 9}{x ^{2} + 2 x + 2} d x

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

\int \frac{- 2 x - 9}{x ^{2} + 2 x + 2} d x = - ln x^{2} + 2 x + 2 - 7 arctan (x + 1)

= 5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1)

Add a constant to the solution

= 5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C

\int 5 - x d x

\int \frac{tan ( x )}{sec ^{3} ( x )} d x

\int \frac{x ^{4} + x ^{2} - 1}{x ^{3} + x} d x

\int sin (2 x) sec (2 x) d x

What is the integral of (3x^2-x+1)/((x+1)(x^2+2x+2)) ?
The integral of (3x^2-x+1)/((x+1)(x^2+2x+2)) is 5ln|x+1|-ln|x^2+2x+2|-7arctan(x+1)+C