Let \( u = 2x - 1 \), so \( du = 2\,dx \) and \( dx = \dfrac{1}{2}du \).

When \( x = 2 \), \( u = 3 \); when \( x = 3 \), \( u = 5 \).

\[ \int_2^3 \frac{1}{2x-1}\,dx = \int_3^5 \frac{1}{u} \cdot \frac{1}{2}\,du = \frac{1}{2}\int_3^5 \frac{1}{u}\,du = \frac{1}{2}\bigl[\ln|u|\bigr]_3^5 = \frac{1}{2}\ln\frac{5}{3}. \]

Numerically, \( \dfrac{1}{2}\ln\dfrac{5}{3} \approx 0.255 \) (3 s.f.).

Answer: 0.255.

Let \( u = 2x - 1 \), so \( du = 2\,dx \) and \( dx = \dfrac{1}{2}du \).

When \( x = 5 \), \( u = 9 \); when \( x = 7 \), \( u = 13 \).

\[ \int_5^7 \frac{2}{2x-1}\,dx = \int_9^{13} \frac{2}{u} \cdot \frac{1}{2}\,du = \int_9^{13} \frac{1}{u}\,du = [\ln|u|]_9^{13} = \ln\frac{13}{9}. \]

Numerically, \( \ln\dfrac{13}{9} \approx 0.368 \) (3 s.f.).

Answer: 0.368.

Let \( u = 7x + 3 \), so \( du = 7\,dx \) and \( dx = \dfrac{1}{7}du \).

When \( x = 0 \), \( u = 3 \); when \( x = 1 \), \( u = 10 \).

\[ \int_0^1 \frac{7}{7x+3}\,dx = \int_3^{10} \frac{7}{u} \cdot \frac{1}{7}\,du = \int_3^{10} \frac{1}{u}\,du = [\ln|u|]_3^{10} = \ln\frac{10}{3}. \]

Numerically, \( \ln\dfrac{10}{3} \approx 1.20 \) (3 s.f.).

Answer: 1.20.

Let \( u = 5 - 2x \), so \( du = -2\,dx \) and \( dx = -\dfrac{1}{2}du \).

When \( x = 0 \), \( u = 5 \); when \( x = 1 \), \( u = 3 \).

\[ \int_0^1 \frac{4}{5-2x}\,dx = \int_5^3 \frac{4}{u} \cdot \Bigl(-\frac{1}{2}\Bigr)du = -2 \int_5^3 \frac{1}{u}\,du = -2[\ln|u|]_5^3 = 2\ln\frac{5}{3}. \]

Numerically, \( 2\ln\dfrac{5}{3} \approx 1.02 \) (3 s.f.).

Answer: 1.02.

Let \( u = 4 - 2x \), so \( du = -2\,dx \) and \( dx = -\dfrac{1}{2}du \).

When \( x = 0 \), \( u = 4 \); when \( x = 1 \), \( u = 2 \).

\[ \int_0^1 \frac{9}{4-2x}\,dx = \int_4^2 \frac{9}{u} \cdot \Bigl(-\frac{1}{2}\Bigr)du = -\frac{9}{2}\int_4^2 \frac{1}{u}\,du = -\frac{9}{2}[\ln|u|]_4^2 = \frac{9}{2}\ln 2. \]

Numerically, \( \dfrac{9}{2}\ln 2 \approx 3.12 \) (3 s.f.).

Answer: 3.12.