Question

can solve this definite integral sum please send it​

Answer 1

Answer:

Step-by-step explanation:

$\bf \int\limits^1_0 {\frac{1}{2x-3} } \, dx \\\\=\frac12 \int\limits^1_0 {\frac{2}{2x-3} } \, dx\\\\=\frac 12 \log (2x-3) \Biggr|_{0}^{1}\\\\=\frac12 (\log ( 2\times1-3)-\log ( 2\times0-3))\\\\=\frac12(\log (-1) -\log (-3))\\\\=\frac12\log \frac {-1}{-3}\\\\=\frac12 \log \frac {1}{3}$