What is the integral of (x^2-4x+3)/(x(x+1)^2) ?

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

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integral of (x^2-4x+3)/(x(x+1)^2)

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

- 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1} + 3 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

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

= ln ∣ x + 1 ∣ + \frac{1}{x + 1} - (- \frac{4}{x + 1}) + 3 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Simplify

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

= - 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1} + 3 ln ∣ x ∣

Add a constant to the solution

= - 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1} + 3 ln ∣ x ∣ + C

\int x^{5} e^{x^{6} + 1} d x

∣ x ∣

\frac{d}{d x} (\frac{1}{6} (e^{3 x} + e^{- 3 x}))

\frac{d x}{d t} = 3 x (5 - x)

\frac{\partial}{\partial x} (\frac{x}{x ^{2} - 2 x})

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