What is the integral of (x+1)/(2x^4+9x^3+8x^2+18x+8) ?

The integral of (x+1)/(2x^4+9x^3+8x^2+18x+8) is

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integral of (x+1)/(2x^4+9x^3+8x^2+18x+8)

\int \frac{x + 1}{2 x ^{4} + 9 x ^{3} + 8 x ^{2} + 18 x + 8} d x

\frac{1}{42} ln ∣ x + 4 ∣ + \frac{2}{63} ln ∣ 2 x + 1 ∣ + \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{x + 1}{2 x ^{4} + 9 x ^{3} + 8 x ^{2} + 18 x + 8} d x

Take the partial fraction of

\frac{x + 1}{2 x ^{4} + 9 x ^{3} + 8 x ^{2} + 18 x + 8} : \frac{1}{42 ( x + 4 )} + \frac{4}{63 ( 2 x + 1 )} + \frac{- x + 2}{18 ( x ^{2} + 2 )}

= \int \frac{1}{42 ( x + 4 )} + \frac{4}{63 ( 2 x + 1 )} + \frac{- x + 2}{18 ( x ^{2} + 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{1}{42 ( x + 4 )} d x + \int \frac{4}{63 ( 2 x + 1 )} d x + \int \frac{- x + 2}{18 ( x ^{2} + 2 )} d x

\int \frac{1}{42 ( x + 4 )} d x = \frac{1}{42} ln ∣ x + 4 ∣

\int \frac{4}{63 ( 2 x + 1 )} d x = \frac{2}{63} ln ∣ 2 x + 1 ∣

\int \frac{- x + 2}{18 ( x ^{2} + 2 )} d x = \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2}))

= \frac{1}{42} ln ∣ x + 4 ∣ + \frac{2}{63} ln ∣ 2 x + 1 ∣ + \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{42} ln ∣ x + 4 ∣ + \frac{2}{63} ln ∣ 2 x + 1 ∣ + \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2})) + C

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What is the integral of (x+1)/(2x^4+9x^3+8x^2+18x+8) ?
The integral of (x+1)/(2x^4+9x^3+8x^2+18x+8) is