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

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

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

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

\frac{10}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{10}{21} ln 4 x^{2} - 4 x + 4 + \frac{20}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{1}{21} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 4 x + 1}{( x ^{2} - x + 1 ) ( x + 4 )} : \frac{20 x + 5}{21 ( x ^{2} - x + 1 )} + \frac{1}{21 ( x + 4 )}

= \int \frac{20 x + 5}{21 ( x ^{2} - x + 1 )} + \frac{1}{21 ( x + 4 )} 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{20 x + 5}{21 ( x ^{2} - x + 1 )} d x + \int \frac{1}{21 ( x + 4 )} d x

\int \frac{20 x + 5}{21 ( x ^{2} - x + 1 )} d x = \frac{1}{21 3} (10 arctan (\frac{1}{3} (2 x - 1)) + 103 ln 4 x^{2} - 4 x + 4 + 20 arctan (\frac{2 x - 1}{3}))

\int \frac{1}{21 ( x + 4 )} d x = \frac{1}{21} ln ∣ x + 4 ∣

= \frac{1}{21 3} (10 arctan (\frac{1}{3} (2 x - 1)) + 103 ln 4 x^{2} - 4 x + 4 + 20 arctan (\frac{2 x - 1}{3})) + \frac{1}{21} ln ∣ x + 4 ∣

Simplify

\frac{1}{21 3} (10 arctan (\frac{1}{3} (2 x - 1)) + 103 ln 4 x^{2} - 4 x + 4 + 20 arctan (\frac{2 x - 1}{3})) + \frac{1}{21} ln ∣ x + 4 ∣ : \frac{10}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{10}{21} ln 4 x^{2} - 4 x + 4 + \frac{20}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{1}{21} ln ∣ x + 4 ∣

= \frac{10}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{10}{21} ln 4 x^{2} - 4 x + 4 + \frac{20}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{1}{21} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{10}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{10}{21} ln 4 x^{2} - 4 x + 4 + \frac{20}{21 3} arctan (\frac{1}{3} (2 x - 1)) + \frac{1}{21} ln ∣ x + 4 ∣ + C

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