# Find the definite integral int_0^1dx/(2x-3)?

Oct 12, 2016

${\int}_{0}^{1} \frac{\mathrm{dx}}{2 x - 3} = - 0.5493$

#### Explanation:

Let $u = 2 x - 3$, hence $\mathrm{du} = 2 \mathrm{dx}$ and

$\int \frac{\mathrm{dx}}{2 x - 3} = \frac{1}{2} \int \frac{\mathrm{du}}{u} = \frac{1}{2} \ln u = \frac{1}{2} \ln | 2 x - 3 |$

Hence, ${\int}_{0}^{1} \frac{\mathrm{dx}}{2 x - 3}$

= $\frac{1}{2} \ln | 2 x - 3 {|}_{0}^{1}$

= $\frac{1}{2} \ln | - 1 | - \frac{1}{2} \ln | - 3 |$

= $\frac{1}{2} \ln 1 - \frac{1}{2} \ln 3$

= $0 - 0.5493$

= $- 0.5493$