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.