$\int_2^3 \frac{d x}{x^2-x}$ is equal to
$\int_2^3 \frac{d x}{x^2-x}$ is equal to
- $\log \frac{2}{3}$
- $\log \frac{4}{3}$
- $\log \frac{8}{3}$
- $\log \frac{1}{4}$
Solution
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}
$
Asked in: AP EAMCET 2002
Practice more Definite Integration questions on Aicharya