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

The integral of (x+1)/((x-4)(x^2+4x+8)) is 1/8 ln|x-4|-1/16 (ln| 1/4 x^2+x+2|-2arctan(1/2 x+1))+C

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integral of (x+1)/((x-4)(x^2+4x+8))

\int \frac{x + 1}{( x - 4 ) ( x ^{2} + 4 x + 8 )} d x

\frac{1}{8} ln ∣ x - 4 ∣ - \frac{1}{16} (ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1)) + C

Solution steps

\int \frac{x + 1}{( x - 4 ) ( x ^{2} + 4 x + 8 )} d x

Take the partial fraction of

\frac{x + 1}{( x - 4 ) ( x ^{2} + 4 x + 8 )} : \frac{1}{8 ( x - 4 )} - \frac{x}{8 ( x ^{2} + 4 x + 8 )}

= \int \frac{1}{8 ( x - 4 )} - \frac{x}{8 ( x ^{2} + 4 x + 8 )} 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}{8 ( x - 4 )} d x - \int \frac{x}{8 ( x ^{2} + 4 x + 8 )} d x

\int \frac{1}{8 ( x - 4 )} d x = \frac{1}{8} ln ∣ x - 4 ∣

\int \frac{x}{8 ( x ^{2} + 4 x + 8 )} d x = \frac{1}{8} (\frac{1}{2} ln \frac{1}{4} (x + 2)^{2} + 1 - arctan (\frac{1}{2} (x + 2)))

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

Simplify

\frac{1}{8} ln ∣ x - 4 ∣ - \frac{1}{8} (\frac{1}{2} ln \frac{1}{4} (x + 2)^{2} + 1 - arctan (\frac{1}{2} (x + 2))) : \frac{1}{8} ln ∣ x - 4 ∣ - \frac{1}{16} (ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1))

= \frac{1}{8} ln ∣ x - 4 ∣ - \frac{1}{16} (ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1))

Add a constant to the solution

= \frac{1}{8} ln ∣ x - 4 ∣ - \frac{1}{16} (ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1)) + C

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What is the integral of (x+1)/((x-4)(x^2+4x+8)) ?
The integral of (x+1)/((x-4)(x^2+4x+8)) is 1/8 ln|x-4|-1/16 (ln| 1/4 x^2+x+2|-2arctan(1/2 x+1))+C