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

The integral of (x+4)/(x^3-2x^2+x) is x/(x^2-2x+1)-6/(x-1)-1/((x-1)^2)+4ln|x|-4ln|x-1|+C

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

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

\frac{x}{x ^{2} - 2 x + 1} - \frac{6}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 4 ln ∣ x ∣ - 4 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

\int \frac{x}{x ^{3} - 2 x ^{2} + x} d x = \frac{x}{x ^{2} - 2 x + 1} - \frac{2}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

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

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

Simplify

\frac{x}{x ^{2} - 2 x + 1} - \frac{2}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 4 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1}) : \frac{x}{x ^{2} - 2 x + 1} - \frac{6}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 4 ln ∣ x ∣ - 4 ln ∣ x - 1 ∣

= \frac{x}{x ^{2} - 2 x + 1} - \frac{6}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 4 ln ∣ x ∣ - 4 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{x}{x ^{2} - 2 x + 1} - \frac{6}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 4 ln ∣ x ∣ - 4 ln ∣ x - 1 ∣ + C

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What is the integral of (x+4)/(x^3-2x^2+x) ?
The integral of (x+4)/(x^3-2x^2+x) is x/(x^2-2x+1)-6/(x-1)-1/((x-1)^2)+4ln|x|-4ln|x-1|+C