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

The integral of 4/(x^3-4x^2+8x) is 4(1/8 ln|x|+1/16 (-ln|x^2-4x+8|+2arctan((x-2)/2)))+C

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integral of 4/(x^3-4x^2+8x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{8 x} + \frac{- x + 4}{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

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

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

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

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

Add a constant to the solution

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

What is the integral of 4/(x^3-4x^2+8x) ?
The integral of 4/(x^3-4x^2+8x) is 4(1/8 ln|x|+1/16 (-ln|x^2-4x+8|+2arctan((x-2)/2)))+C