Answer to Question #233062 in Calculus for moe

Question

Consider the integral $\displaystyle{\int \frac{2}{x^2-2x}dx}=\displaystyle{\ln{\frac{x-a}{x}}+C}$

 Which of the following values represent the constant a ?

 

1. 1
2. 7
3. 4
4. 2

Answer 1

Solve the integral by partial fractions

$$\dfrac{2}{x^2-2x}=\dfrac{2}{x(x-2)}=\dfrac{A}{x}+\dfrac{B}{x-2}$$

The sum is added up:

$$\dfrac{2}{x^2-2x}=\dfrac{A(x-2)+Bx}{x(x-2)}$$

Since the two fractions have the same denominator, the numerators must be equal:

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

To calculate the values of $A$ , $B$ , we give $x$ the values that cancel out the denominator

$$x=0\Rightarrow 2=-2A\Rightarrow A=-1$$

$$x=2\Rightarrow 2=2B\Rightarrow B=1$$

Integrals of simple fractions are calculated:

$$\int \dfrac{2}{x^2-2x}\, dx=-\int \dfrac1x\, dx+\int \dfrac{1}{x-2}\, dx$$

$$=-\ln |x|+\ln |x-2|+C=\ln \left| \dfrac{x-2}{x}\right|+C$$

Comparing with the expression of the statement we see that $\boxed{a=2}.$