What is the integral of (8x+5)/((x^2+4x+8)(x-1)) ?

The integral of (8x+5)/((x^2+4x+8)(x-1)) is 1/2 (-ln| 1/4 x^2+x+2|+5arctan(1/2 x+1))+ln|x-1|+C

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integral of (8x+5)/((x^2+4x+8)(x-1))

\int \frac{8 x + 5}{( x ^{2} + 4 x + 8 ) ( x - 1 )} d x

\frac{1}{2} (- ln \frac{1}{4} x^{2} + x + 2 + 5 arctan (\frac{1}{2} x + 1)) + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{8 x + 5}{( x ^{2} + 4 x + 8 ) ( x - 1 )} d x

Take the partial fraction of

\frac{8 x + 5}{( x ^{2} + 4 x + 8 ) ( x - 1 )} : \frac{- x + 3}{x ^{2} + 4 x + 8} + \frac{1}{x - 1}

= \int \frac{- x + 3}{x ^{2} + 4 x + 8} + \frac{1}{x - 1} 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{- x + 3}{x ^{2} + 4 x + 8} d x + \int \frac{1}{x - 1} d x

\int \frac{- x + 3}{x ^{2} + 4 x + 8} d x = \frac{1}{2} (- ln \frac{1}{4} (x + 2)^{2} + 1 + 5 arctan (\frac{1}{2} (x + 2)))

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= \frac{1}{2} (- ln \frac{1}{4} (x + 2)^{2} + 1 + 5 arctan (\frac{1}{2} (x + 2))) + ln ∣ x - 1 ∣

Simplify

\frac{1}{2} (- ln \frac{1}{4} (x + 2)^{2} + 1 + 5 arctan (\frac{1}{2} (x + 2))) + ln ∣ x - 1 ∣ : \frac{1}{2} (- ln \frac{1}{4} x^{2} + x + 2 + 5 arctan (\frac{1}{2} x + 1)) + ln ∣ x - 1 ∣

= \frac{1}{2} (- ln \frac{1}{4} x^{2} + x + 2 + 5 arctan (\frac{1}{2} x + 1)) + ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{2} (- ln \frac{1}{4} x^{2} + x + 2 + 5 arctan (\frac{1}{2} x + 1)) + ln ∣ x - 1 ∣ + C

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What is the integral of (8x+5)/((x^2+4x+8)(x-1)) ?
The integral of (8x+5)/((x^2+4x+8)(x-1)) is 1/2 (-ln| 1/4 x^2+x+2|+5arctan(1/2 x+1))+ln|x-1|+C