What is the integral of (7x^2-4x+4)/((x+1)(x^2-x+1)) ?

The integral of (7x^2-4x+4)/((x+1)(x^2-x+1)) is

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integral of (7x^2-4x+4)/((x+1)(x^2-x+1))

\int \frac{7 x ^{2} - 4 x + 4}{( x + 1 ) ( x ^{2} - x + 1 )} d x

\frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) + \frac{8}{3} ln ∣ x + 1 ∣ - \frac{2}{3} ln 4 x^{2} - 4 x + 4 - \frac{4}{3} arctan (\frac{1}{3} (2 x - 1)) + \frac{2}{3} (23 arctan (\frac{2 x - 1}{3}) - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3}) + C

Solution steps

\int \frac{7 x ^{2} - 4 x + 4}{( x + 1 ) ( x ^{2} - x + 1 )} d x

Expand

\frac{7 x ^{2} - 4 x + 4}{( x + 1 ) ( x ^{2} - x + 1 )} : \frac{7 x ^{2}}{( x + 1 ) ( x ^{2} - x + 1 )} - \frac{4 x}{( x + 1 ) ( x ^{2} - x + 1 )} + \frac{4}{( x + 1 ) ( 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{7 x ^{2}}{( x + 1 ) ( x ^{2} - x + 1 )} d x - \int \frac{4 x}{( x + 1 ) ( x ^{2} - x + 1 )} d x + \int \frac{4}{( x + 1 ) ( x ^{2} - x + 1 )} d x

\int \frac{7 x ^{2}}{( x + 1 ) ( x ^{2} - x + 1 )} d x = \frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1)

\int \frac{4 x}{( x + 1 ) ( x ^{2} - x + 1 )} d x = 4 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3})))

\int \frac{4}{( x + 1 ) ( x ^{2} - x + 1 )} d x = 4 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

= \frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) - 4 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))) + 4 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

Simplify

\frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) - 4 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))) + 4 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))) : \frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) + \frac{8}{3} ln ∣ x + 1 ∣ - \frac{2}{3} ln 4 x^{2} - 4 x + 4 - \frac{4}{3} arctan (\frac{1}{3} (2 x - 1)) + \frac{2}{3} (23 arctan (\frac{2 x - 1}{3}) - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3})

= \frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) + \frac{8}{3} ln ∣ x + 1 ∣ - \frac{2}{3} ln 4 x^{2} - 4 x + 4 - \frac{4}{3} arctan (\frac{1}{3} (2 x - 1)) + \frac{2}{3} (23 arctan (\frac{2 x - 1}{3}) - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3})

Add a constant to the solution

= \frac{7}{3} ln (∣ x + 1 ∣ x^{2} - x + 1) + \frac{8}{3} ln ∣ x + 1 ∣ - \frac{2}{3} ln 4 x^{2} - 4 x + 4 - \frac{4}{3} arctan (\frac{1}{3} (2 x - 1)) + \frac{2}{3} (23 arctan (\frac{2 x - 1}{3}) - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3}) + C

(3 + csc (2 t))^{- 6}

f (x) = - 2 - cos (x), at x = - \frac{π}{4}

\frac{1}{( s - 1 )}

f (x) = x^{2} - 7, (2, - 3)

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

What is the integral of (7x^2-4x+4)/((x+1)(x^2-x+1)) ?
The integral of (7x^2-4x+4)/((x+1)(x^2-x+1)) is