Evaluate # int \ 1/(x^2-2x) \ dx #?

Question

Charles Anctil · Accepted Answer

# int \ 1/(x^2-2x) \ dx = 1/2 ln |(A(x-2))/x| #

We want to find: # int \ 1/(x^2-2x) \ dx # If we examine the integrand we can decompose into partial fractions, as follows: # 1/(x^2-2x) = 1/(x(x-2)) # # " " = A/x + B/(x-2) # # " " = ( A(x-2) + Bx ) /(x(x-2)) # And so: # 1 -= A(x-2) + Bx # Put: # x = 0=> 1 = -2A => A = -1/2 # # x = 2=> 1 = 2B \ \ \ \ \=> B = 1/2 # Therefore: # int \ 1/(x^2-2x) \ dx = int \ (1/2)/(x-2) - (1/2)/x \ dx# # " " = 1/2 ln |x-2| - 1/2ln|x| + C# # " " = 1/2 ln |(x-2)/x| + C# # " " = 1/2 ln |(A(x-2))/x| #

Charles Anctil · Answer

undefined To evaluate the integral $\int \frac{1}{x^2 - 2x} \, dx$, we first factor the denominator:

$$x^2 - 2x = x(x - 2)$$

Now, we use partial fraction decomposition to break down the fraction:

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

Multiplying both sides by $x(x - 2)$, we get:

$$1 = A(x - 2) + Bx$$

Expanding and collecting like terms:

$$1 = (A + B)x - 2A$$

This equation must hold for all values of $x$, so the coefficients must be equal:

$$A + B = 0$$ 
$$-2A = 1$$

Solving this system of equations, we find $A = -\frac{1}{2}$ and $B = \frac{1}{2}$.

Now, we can rewrite the integral as:

$$\int \left(-\frac{1}{2x} + \frac{1}{2(x - 2)}\right) \, dx$$

Which integrates to:

$$-\frac{1}{2} \ln |x| + \frac{1}{2} \ln |x - 2| + C$$

Where $C$ is the constant of integration.