How do you integrate #(x^2+2x-1) / (x^3 - x)#?

Question

Christopher Allers · Accepted Answer

#int(x^2+2x-1)/(x^3-x)dx=lnx-ln(x+1)+ln(x-1)+c# As #(x^2+2x-1)/(x^3-x)=(x^2+2x-1)/(x(x+1)(x-1))#, let us convert them into partial fractions #(x^2+2x-1)/(x^3-x)hArrA/x+B/(x+1)+C/(x-1)# or #(x^2+2x-1)/(x^3-x)=(A(x^2-1)+B(x^2-x)+C(x^2+x))/(x(x+1)(x-1))# or #(x^2+2x-1)/(x^3-x)=((A+B+C)x^2+(-B+C)x-A)/(x(x+1)(x-1))# i.e. #A+B+C=1#, #-B+C=2# and #A=1#, which gives #C=1# and #B=-1# and hence #(x^2+2x-1)/(x^3-x)=1/x-1/(x+1)+1/(x-1)# and #int(x^2+2x-1)/(x^3-x)dx=int[1/x-1/(x+1)+1/(x-1)]dx# = #lnx-ln(x+1)+ln(x-1)+c#

Luke Alvarez · Answer

undefined To integrate the function $ \frac{x^2 + 2x - 1}{x^3 - x} $, we can perform polynomial long division first to simplify the expression. After that, we can decompose the resulting fraction into partial fractions if necessary. However, in this case, since the degree of the numerator is less than the degree of the denominator by one, we can apply a straightforward decomposition by recognizing the form of the antiderivative.

The decomposition would be as follows:

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

Now, we can rewrite the original expression as a sum of simpler fractions:

$$ \frac{x^2 + 2x - 1}{x(x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1} $$

Next, we can find the values of $ A $, $ B $, and $ C $ by equating coefficients. After finding $ A $, $ B $, and $ C $, we integrate each term separately.

Finally, integrate each term separately and sum them up to get the final result.