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.

Let \( u = 5 - x \), so \( du = -dx \) and \( dx = -du \).

When \( x = 0 \), \( u = 5 \); when \( x = 2 \), \( u = 3 \).

\[ \int_0^2 \frac{-7}{5-x}\,dx = \int_5^3 \frac{-7}{u}(-du) = 7\int_5^3 \frac{1}{u}\,du = 7[\ln|u|]_5^3 = 7\ln\frac{3}{5}. \]

Since \( \ln\dfrac{3}{5} < 0 \), the value is negative.

Numerically, \( 7\ln\dfrac{3}{5} \approx -3.58 \) (3 s.f.).

Answer: -3.58.

Let \( u = x - 7 \), so \( du = dx \).

When \( x = 2 \), \( u = -5 \); when \( x = 4 \), \( u = -3 \).

\[ \int_2^4 \frac{2}{x-7}\,dx = 2\int_{-5}^{-3} \frac{1}{u}\,du = 2[\ln|u|]_{-5}^{-3} = 2(\ln 3 - \ln 5) = 2\ln\frac{3}{5}. \]

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

Answer: -1.02.

Let \( u = 7 - 2x \), so \( du = -2\,dx \) and \( dx = -\dfrac{1}{2}du \).

When \( x = -6 \), \( u = 19 \); when \( x = -4 \), \( u = 15 \).

\[ \int_{-6}^{-4} \frac{-4}{7-2x}\,dx = \int_{19}^{15} \frac{-4}{u}\cdot\Bigl(-\frac{1}{2}\Bigr)du = 2\int_{19}^{15} \frac{1}{u}\,du = 2[\ln|u|]_{19}^{15} = 2(\ln 15 - \ln 19) = 2\ln\frac{15}{19}. \]

Numerically, \( 2\ln\dfrac{15}{19} \approx -0.473 \) (3 s.f.).

Answer: -0.473.

Let \( u = 5 - 3x \), so \( du = -3\,dx \) and \( dx = -\dfrac{1}{3}du \).

When \( x = -1 \), \( u = 8 \); when \( x = 1 \), \( u = 2 \).

\[ \int_{-1}^{1} \frac{6}{5-3x}\,dx = \int_8^2 \frac{6}{u}\cdot\Bigl(-\frac{1}{3}\Bigr)du = -2\int_8^2 \frac{1}{u}\,du = -2[\ln|u|]_8^2 = 2(\ln 8 - \ln 2) = 2\ln 4 = \ln 16. \]

Numerically, \( \ln 16 \approx 2.77 \) (3 s.f.).

Answer: 2.77.