Question

$\int\frac{dx}{x^2-x}=$

• A. $\log\frac{|x|}{|x-1|}+C$
• B. $\frac{-1}{x}+\log|x-1|+C$
• C. $x\log|x-1|+C$
• D. $\log\frac{|x-1|}{|x|}+C$
• E. $-x\log|x-1|+C$

Answer 1

The Correct Option is D

We are tasked with evaluating the integral: $$\int \frac{dx}{x^2 - x}$$ First, factor the denominator: $$x^2 - x = x(x - 1)$$ So, the integral becomes: $$\int \frac{dx}{x(x - 1)}$$ Next, use partial fraction decomposition to rewrite the integrand: $$\frac{1}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1}$$ Multiply both sides by $x(x - 1)$: $$1 = A(x - 1) + Bx$$ Expanding: $$1 = A(x) - A + Bx$$ $$1 = (A + B)x - A$$ Equating the coefficients of $x$ and the constant term gives: $$A + B = 0 \quad \text{and} \quad -A = 1$$ Solving these equations: $$A = -1, \quad B = 1$$ Thus, the decomposition is: $$\frac{1}{x(x - 1)} = \frac{-1}{x} + \frac{1}{x - 1}$$ Now, integrate term by term: $$\int \frac{dx}{x^2 - x} = \int \left( \frac{-1}{x} + \frac{1}{x - 1} \right) dx$$ $$= -\log |x| + \log |x - 1| + C$$ $$= \log \left| \frac{x - 1}{x} \right| + C$$

The correct option is (D) : $\log\frac{|x-1|}{|x|}+C$