Home > Calculus > Integrals of Rational Functions

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

Answer 1

Charles Arocha

#int (x^2+2x-1)/(x^3-x) dx = ln abs(x) + ln abs(x-1) - ln abs(x+1) + C#

#(x^2+2x-1)/(x^3-x) = (x^2+2x-1)/(x(x-1)(x+1)) = a/x + b/(x-1) + c/(x+1)#

Use Heaviside's cover up method to find #a, b, c#:

#a = ((color(blue)(0))^2+2(color(blue)(0))-1)/(((color(blue)(0))-1)((color(blue)(0))+1)) = (-1)/((-1)(1)) = 1#

#b = ((color(blue)(1))^2+2(color(blue)(1))-1)/((color(blue)(1))((color(blue)(1))+1)) = (2)/((1)(2)) = 1#

#c = ((color(blue)(-1))^2+2(color(blue)(-1))-1)/((color(blue)(-1))((color(blue)(-1))-1)) = (-2)/((-1)(-2)) = -1#

So:

#int (x^2+2x-1)/(x^3-x) dx = int 1/x + 1/(x-1) -1/(x+1) dx#

#color(white)(int (x^2+2x-1)/(x^3-x) dx) = ln abs(x) + ln abs(x-1) - ln abs(x+1) + C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Christopher Agresta

To integrate ( \frac{x^2 + 2x - 1}{x^3 - x} , dx ), you can perform polynomial long division to rewrite the expression. Once you have simplified the integrand using polynomial long division, you can decompose it into partial fractions. Then, integrate each partial fraction individually. Finally, you can combine the results to obtain the integral of the original expression.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.