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

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

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

integral

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

x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 + \frac{1}{2 3} (ln \frac{1}{3} (x - 2) + 1 - ln \frac{1}{3} (x - 2) - 1) + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{2} - 4 x + 1} d x = x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3

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

= x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 - (- \frac{1}{2 3} (ln \frac{x - 2}{3} + 1 - ln \frac{x - 2}{3} - 1))

Simplify

x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 - (- \frac{1}{2 3} (ln \frac{x - 2}{3} + 1 - ln \frac{x - 2}{3} - 1)) : x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 + \frac{1}{2 3} (ln \frac{1}{3} (x - 2) + 1 - ln \frac{1}{3} (x - 2) - 1)

= x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 + \frac{1}{2 3} (ln \frac{1}{3} (x - 2) + 1 - ln \frac{1}{3} (x - 2) - 1)

Add a constant to the solution

= x - 2 - \frac{- 4 3 + 7}{2 3} ln x - 2 + 3 + \frac{4 3 + 7}{2 3} ln x - 2 - 3 + \frac{1}{2 3} (ln \frac{1}{3} (x - 2) + 1 - ln \frac{1}{3} (x - 2) - 1) + C

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

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

16 - x^{2}

x^{2} \cdot (e^{x})^{2}

e^{x} - 2^{x}

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