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) : \(\log\frac{|x-1|}{|x|}+C\)

Answer 2

We want to evaluate the integral: \[\int \frac{dx}{x^2 - x}\]

First, we factor the denominator: \[x^2 - x = x(x - 1)\]

Now, we use partial fraction decomposition: \[\frac{1}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1}\]

Multiply both sides by \(x(x - 1)\): \[1 = A(x - 1) + Bx\]

To solve for \(A\) and \(B\), we can use convenient values of \(x\). If \(x = 0\): \[1 = A(0 - 1) + B(0) = -A\] \[A = -1\]

If \(x = 1\): \[1 = A(1 - 1) + B(1) = B\] \[B = 1\]

So, our decomposition is: \[\frac{1}{x(x - 1)} = \frac{-1}{x} + \frac{1}{x - 1}\]

Now, we can integrate: \[\int \frac{dx}{x^2 - x} = \int \left(\frac{-1}{x} + \frac{1}{x - 1}\right) dx = -\int \frac{1}{x} dx + \int \frac{1}{x - 1} dx\]

The integrals are: \[-\int \frac{1}{x} dx = -\ln|x| + C_1\] \[\int \frac{1}{x - 1} dx = \ln|x - 1| + C_2\]

Adding these, we get: \[-\ln|x| + \ln|x - 1| + C = \ln|x - 1| - \ln|x| + C\]

Using the property \(\ln a - \ln b = \ln \frac{a}{b}\): \[\ln \left|\frac{x - 1}{x}\right| + C\]

Therefore, the integral is: \[\int \frac{dx}{x^2 - x} = \ln \left|\frac{x - 1}{x}\right| + C\]