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$\int_2^3 \frac{d x}{x^2-x}$ is equal to
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1836 Upvotes
Verified Answer
The correct answer is:
$\log \frac{4}{3}$
We have,
$$
\begin{aligned}
\int_2^3 \frac{d x}{x^2-x} & =\int_2^3 \frac{1}{x(x-1)} d x \\
& =\int_2^3\left[-\frac{1}{x}+\frac{1}{x-1}\right] d x \\
& =\left[\log \frac{x-1}{x}\right]_2^3=\log \frac{2}{3}-\log \frac{1}{2}=\log \frac{4}{3}
\end{aligned}
$$
$$
\begin{aligned}
\int_2^3 \frac{d x}{x^2-x} & =\int_2^3 \frac{1}{x(x-1)} d x \\
& =\int_2^3\left[-\frac{1}{x}+\frac{1}{x-1}\right] d x \\
& =\left[\log \frac{x-1}{x}\right]_2^3=\log \frac{2}{3}-\log \frac{1}{2}=\log \frac{4}{3}
\end{aligned}
$$
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