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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

x \cdot (1 - x)

\frac{\partial}{\partial y} (sin (x y))

\int (a + x)^{2} d x

\frac{d}{d x} (\frac{( x ^{2} - 1 ) ^{3}}{4})

n = 0 \sum \infty \frac{e ^{n}}{n ^{n}}

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