First decompose into partial fractions:

\[ \frac{1}{x(x-1)^2} = \frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}. \]

So

\[ \int_{2}^{3} \frac{1}{x(x-1)^2}dx = \int_{2}^{3} \left( \frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2} \right) dx. \]

Integrate term by term:

\[ \int \frac{1}{x}dx = \ln|x|,\quad \int \frac{1}{x-1}dx = \ln|x-1|,\quad \int \frac{1}{(x-1)^2}dx = -\frac{1}{x-1}. \]

Hence

\[ \int \frac{1}{x(x-1)^2}dx = \ln\left|\frac{x}{x-1}\right| - \frac{1}{x-1} + C. \]

Now evaluate from 2 to 3:

\[ \Bigl[\ln\left|\tfrac{x}{x-1}\right| - \tfrac{1}{x-1}\Bigr]_{2}^{3} = \left(\ln\tfrac{3}{2} - \tfrac{1}{2}\right) - \left(\ln 2 - 1\right) = \ln\tfrac{3}{4} + \tfrac{1}{2}. \]

Numerically, this is approximately \( 0.212 \) (3 s.f.).

Answer: 0.212.

Partial fractions:

\[ \frac{1}{x(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}. \]

So

\[ \int_{1}^{2} \frac{1}{x(x+1)^2}dx = \int_{1}^{2}\left(\frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}\right)dx. \]

Integrals:

\[ \int \frac{1}{x}dx = \ln|x|,\; \int \frac{1}{x+1}dx = \ln|x+1|,\; \int \frac{1}{(x+1)^2}dx = -\frac{1}{x+1}. \]

Thus

\[ \int \frac{1}{x(x+1)^2}dx = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x+1} + C. \]

Evaluate from 1 to 2:

\[ \Bigl[\ln\tfrac{x}{x+1} + \tfrac{1}{x+1}\Bigr]_1^2 = \left(\ln\tfrac{2}{3} + \tfrac{1}{3}\right) - \left(\ln\tfrac{1}{2} + \tfrac{1}{2}\right) = -\tfrac{1}{6} + \ln\tfrac{4}{3}. \]

Numerically, this is approximately \( 0.121 \) (3 s.f.).

Answer: 0.121.

Partial fractions:

\[ \frac{8x+24}{(x+1)(x-3)^2} = \frac{1}{x+1} - \frac{1}{x-3} + \frac{12}{(x-3)^2}. \]

So

\[ \int_{0}^{2} \frac{8x+24}{(x+1)(x-3)^2}dx = \int_{0}^{2}\left(\frac{1}{x+1} - \frac{1}{x-3} + \frac{12}{(x-3)^2}\right)dx. \]

Integrate:

\[ \int \frac{1}{x+1}dx = \ln|x+1|,\quad \int \frac{1}{x-3}dx = \ln|x-3|,\quad \int \frac{12}{(x-3)^2}dx = -\frac{12}{x-3}. \]

Hence the antiderivative is

\[ F(x) = \ln|x+1| - \ln|x-3| - \frac{12}{x-3} + C. \]

Evaluate from 0 to 2:

\[ F(2) - F(0) = (\ln 3 - \ln 1 - 12/(-1)) - (\ln 1 - \ln 3 - 12/(-3)) = (\ln 3 + 12) - (-\ln 3 + 4) = 8 + 2\ln 3. \]

Numerically, this is approximately \( 10.2 \) (3 s.f.).

Answer: 10.2.

Partial fractions:

\[ \frac{5}{(3+2x)(1-x)} = \frac{2}{3+2x} + \frac{1}{1-x} = \frac{2}{2x+3} - \frac{1}{x-1}. \]

Integrate:

\[ \int \frac{2}{2x+3}dx = \ln|2x+3|,\quad \int \frac{1}{1-x}dx = -\ln|1-x|. \]

So an antiderivative is

\[ F(x) = \ln|2x+3| - \ln|1-x| + C. \]

Evaluate from \( -0.5 \) to \( 0.5 \):

At \( x = 0.5 \): \( F(0.5) = \ln 4 - \ln 0.5 = \ln 8. \)

At \( x = -0.5 \): \( F(-0.5) = \ln 2 - \ln 1.5 = \ln \tfrac{4}{3}. \)

\[ F(0.5) - F(-0.5) = \ln 8 - \ln \tfrac{4}{3} = \ln 6. \]

\( \ln 6 \approx 1.79 \) (3 s.f.).

Answer: 1.79.

Partial fractions:

\[ \frac{5x+5}{(x-2)(x+3)} = \frac{3}{x-2} + \frac{2}{x+3}. \]

So

\[ \int_{3}^{4} \frac{5x+5}{(x-2)(x+3)}dx = \int_{3}^{4}\left(\frac{3}{x-2} + \frac{2}{x+3}\right)dx = \Bigl[3\ln|x-2| + 2\ln|x+3|\Bigr]_{3}^{4}. \]

Compute:

\[ = (3\ln 2 + 2\ln 7) - (3\ln 1 + 2\ln 6) = \ln\frac{2^3 7^2}{6^2} = \ln\frac{98}{9}. \]

\( \ln \tfrac{98}{9} \approx 2.39 \) (3 s.f.).

Answer: 2.39.

Rewrite by partial fractions:

\[ \frac{4-x}{(x-2)^2} = -\frac{1}{x-2} + \frac{2}{(x-2)^2}. \]

So

\[ \int_{7}^{12} \frac{4-x}{(x-2)^2}dx = \int_{7}^{12}\left(-\frac{1}{x-2} + \frac{2}{(x-2)^2}\right)dx. \]

Integrate:

\[ \int -\frac{1}{x-2}dx = -\ln|x-2|,\quad \int \frac{2}{(x-2)^2}dx = -\frac{2}{x-2}. \]

Thus

\[ F(x) = -\ln|x-2| - \frac{2}{x-2} + C. \]

Evaluate from 7 to 12:

\[ F(12) - F(7) = \left(-\ln 10 - \tfrac{2}{10}\right) - \left(-\ln 5 - \tfrac{2}{5}\right) = (\ln 5 - \ln 2) + \left(\tfrac{2}{5} - \tfrac{1}{5}\right) = -\ln 2 + \tfrac{1}{5}. \]

Numerically, this is \( \approx -0.493 \) (3 s.f.).

Answer: -0.493.

Partial fractions:

\[ \frac{25}{(2x-3)^2(1+x)} = -\frac{2}{2x-3} + \frac{10}{(2x-3)^2} + \frac{1}{1+x}. \]

Integrate each term:

\[ \int -\frac{2}{2x-3}dx = -\ln|2x-3|,\quad \int \frac{10}{(2x-3)^2}dx = -\frac{5}{2x-3},\quad \int \frac{1}{1+x}dx = \ln|1+x|. \]

So an antiderivative is

\[ F(x) = -\ln|2x-3| - \frac{5}{2x-3} + \ln|1+x| + C. \]

Evaluate from 4 to 5:

\[ F(5) - F(4) = (-\ln 7 - 5/7 + \ln 6) - (-\ln 5 - 1 + \ln 5) = \ln\tfrac{6}{7} + \tfrac{2}{7}. \]

Numerically, this is approximately \( 0.132 \) (3 s.f.).

Answer: 0.132.

Factor the denominator:

\[ x^3 - 4x^2 + 4x = x(x-2)^2. \]

Then

\[ \frac{4}{x(x-2)^2} = \frac{1}{x} - \frac{1}{x-2} + \frac{2}{(x-2)^2}. \]

So

\[ \int_{3}^{4} \frac{4}{x^3 - 4x^2 + 4x}dx = \int_{3}^{4}\left(\frac{1}{x} - \frac{1}{x-2} + \frac{2}{(x-2)^2}\right)dx. \]

Antiderivative:

\[ F(x) = \ln|x| - \ln|x-2| - \frac{2}{x-2} + C. \]

Evaluate from 3 to 4:

\[ F(4) - F(3) = (\ln 4 - \ln 2 - 2/2) - (\ln 3 - \ln 1 - 2/1) = \ln\tfrac{2}{3} + 1. \]

Numerically, this is \( \approx 0.595 \) (3 s.f.).

Answer: 0.595.

Partial fractions / algebraic manipulation gives

\[ \frac{x^2 - 2x + 6}{(x+1)(x-2)} = 1 - \frac{3}{x+1} + \frac{2}{x-2}. \]

So

\[ \int_{0}^{1} \frac{x^2 - 2x + 6}{(x+1)(x-2)}dx = \int_0^1\left(1 - \frac{3}{x+1} + \frac{2}{x-2}\right)dx. \]

Antiderivative:

\[ F(x) = x - 3\ln|x+1| + 2\ln|x-2| + C. \]

Evaluate from 0 to 1:

\[ F(1) - F(0) = \bigl(1 - 3\ln 2 + 2\ln 1\bigr) - \bigl(0 - 3\ln 1 + 2\ln 2\bigr) = 1 - 5\ln 2 = 1 - \ln 32. \]

Numerically, this is approximately \( -2.47 \) (3 s.f.).

Answer: -2.47.