Problem 8. \( \int_{1}^{\infty} \frac{1}{2025[x]} \mathrm{d} x \) Problem 9. \( \int_{4}^{5} \frac{2 x+1}{(x-1)(x-2)(x-3)} \mathrm{d} x \) Problem 10. \( \int_{0}^{1} x e^{x^{2}+e^{x^{2}}} \mathrm{~d} x \)

**Problem 8** We wish to evaluate \[ \int_{1}^{\infty} \frac{1}{2025[x]} \, dx, \] where \([x]\) denotes the greatest integer less than or equal to \(x\). Since \([x]\) is constant on any interval \([n,\, n+1)\) for integer \(n\), we break up the integral as follows: \[ \int_{1}^{\infty} \frac{1}{2025[x]} \, dx=\frac{1}{2025}\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{dx}{n}. \] On the interval \(x\in [n,\, n+1)\) we have \([x] = n\) and the length of the interval is 1, so \[ \int_{n}^{n+1}\frac{dx}{n}=\frac{1}{n}. \] Thus, \[ \int_{1}^{\infty} \frac{1}{2025[x]} \, dx=\frac{1}{2025}\sum_{n=1}^{\infty}\frac{1}{n}. \] The series \(\sum_{n=1}^{\infty}\frac{1}{n}\) is the harmonic series, which diverges. Therefore, the integral diverges. --- **Problem 9** We want to compute \[ \int_{4}^{5} \frac{2x+1}{(x-1)(x-2)(x-3)} \, dx. \] **Step 1. Partial Fraction Decomposition** Write \[ \frac{2x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}. \] Multiplying both sides by \((x-1)(x-2)(x-3)\) gives: \[ 2x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2). \] Expand each term: \[ A(x-2)(x-3)=A(x^2-5x+6), \] \[ B(x-1)(x-3)=B(x^2-4x+3), \] \[ C(x-1)(x-2)=C(x^2-3x+2). \] Thus, \[ 2x+1 = (A+B+C)x^2 + (-5A-4B-3C)x + (6A+3B+2C). \] Equate coefficients with the left-hand side (which has 0 as the coefficient of \(x^2\)): \[ \begin{cases} A+B+C=0,\\[1mm] -5A-4B-3C = 2,\\[1mm] 6A+3B+2C = 1. \end{cases} \] From the first equation, \(C = -A-B\). Substitute this into the other two equations: For the \(x\)-coefficient: \[ -5A-4B-3(-A-B) = -5A-4B+3A+3B = -2A-B = 2. \] Thus, \[ -2A - B = 2 \quad \Longrightarrow \quad B = -2A - 2. \] For the constant term: \[ 6A+3B+2(-A-B) = 6A+3B-2A-2B = 4A+B = 1. \] Substitute \(B = -2A-2\): \[ 4A + (-2A-2) = 1 \quad \Longrightarrow \quad 2A - 2 = 1 \quad \Longrightarrow \quad 2A=3, \] so \[ A=\frac{3}{2}. \] Then, \[ B=-2\left(\frac{3}{2}\right)-2 = -3-2 = -5, \] and \[ C=-A-B=-\frac{3}{2}-(-5)= -\frac{3}{2}+5=\frac{7}{2}. \] Thus, the decomposition is: \[ \frac{2x+1}{(x-1)(x-2)(x-3)}=\frac{\frac{3}{2}}{x-1}-\frac{5}{x-2}+\frac{\frac{7}{2}}{x-3}. \] **Step 2. Integration** Now integrate term by term from \(4\) to \(5\): \[ \int_{4}^{5} \frac{2x+1}{(x-1)(x-2)(x-3)} \, dx = \frac{3}{2}\int_{4}^{5}\frac{dx}{x-1} - 5\int_{4}^{5}\frac{dx}{x-2} + \frac{7}{2}\int_{4}^{5}\frac{dx}{x-3}. \] The antiderivative of \(\frac{1}{x-a}\) is \(\ln{|x-a|}\), so: \[ \frac{3}{2}\ln{|x-1|} - 5\ln{|x-2|} + \frac{7}{2}\ln{|x-3|}\Bigg|_{x=4}^{x=5}. \] Evaluate at the upper limit \(x=5\): \[ \begin{aligned} x=5:&\quad \ln{|5-1|}=\ln{4},\quad \ln{|5-2|}=\ln{3},\quad \ln{|5-3|}=\ln{2},\\[1mm] \text{Thus:}\quad &\frac{3}{2}\ln{4}-5\ln{3}+\frac{7}{2}\ln{2}. \end{aligned} \] Evaluate at the lower limit \(x=4\): \[ \begin{aligned} x=4:&\quad \ln{|4-1|}=\ln{3},\quad \ln{|4-2|