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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x - 2} d x = 2 ln ∣ x - 2 ∣

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

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

Add a constant to the solution

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

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