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

y = 4 x^{2}, x = 1, y = 0

\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}

y = e^{\frac{( 3 x ^{2} )}{2}}

\frac{x ^{63} - 1}{x - 1}

\frac{d}{d x} (\frac{3}{3 x + 1})

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