Câu hỏi:

Ta có tính chất của tích phân như sau: $\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$.
Vậy $\int_a^c f(x) dx = -5 + 10 = 5$.

Ta có công thức nguyên hàm cơ bản: $\int \sin x dx = -\cos x + C$.
Vậy, $\int {2026\sin x{\rm{d}}x} = 2026\int \sin x dx = 2026(-\cos x) + C = -2026\cos x + C$.

Ta có: $\int\limits_0^2 {[f(x) - 3x^2 ]{\rm{d}}x} = \int\limits_0^2 {f(x){\rm{d}}x} - \int\limits_0^2 {3x^2{\rm{d}}x} = 4$.
Tính $\int\limits_0^2 {3x^2{\rm{d}}x} = x^3\Big|_0^2 = 2^3 - 0^3 = 8$.
Suy ra $\int\limits_0^2 {f(x){\rm{d}}x} - 8 = 4 \Rightarrow \int\limits_0^2 {f(x){\rm{d}}x} = 12$.

Ta có: $f(x) = 3^{x-1} \cdot 5^{x+1} = 3^x \cdot 3^{-1} \cdot 5^x \cdot 5^1 = \frac{1}{3} \cdot 5 \cdot 3^x \cdot 5^x = \frac{5}{3} \cdot (3 \cdot 5)^x = \frac{5}{3} \cdot 15^x$. Do đó, $\int f(x) dx = \int \frac{5}{3} \cdot 15^x dx = \frac{5}{3} \int 15^x dx = \frac{5}{3} \cdot \frac{15^x}{\ln 15} + C$.

Ta có: $2x^2 - x - 1 = (2x+1)(x-1)$.
Khi đó:
$\frac{3x+1}{2x^2 - x - 1} = \frac{3x+1}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1}$
$3x+1 = A(x-1) + B(2x+1)$
Chọn $x=1$ ta có $4 = 3B \Rightarrow B = \frac{4}{3}$
Chọn $x = -\frac{1}{2}$ ta có $-\frac{1}{2} = -\frac{3}{2}A \Rightarrow A = \frac{1}{3}$
Vậy $\int\limits_2^3 {\frac{{3x + 1}}{{2{x^2} - x - 1}}{\rm{d}}x} = \int\limits_2^3 {\left( {\frac{1}{{3(2x + 1)}} + \frac{4}{{3(x - 1)}}} \right){\rm{d}}x} $
$= \frac{1}{6}\ln (2x+1) \Big|^3_2 + \frac{4}{3} \ln (x-1) \Big|^3_2$
$= \frac{1}{6} (\ln 7 - \ln 5) + \frac{4}{3} (\ln 2 - \ln 1) $
$= \frac{4}{3}\ln 2 - \frac{1}{6}\ln 5 + \frac{1}{6}\ln 7$
Suy ra $a = \frac{4}{3}, b = -\frac{1}{6}, c = \frac{1}{6}$
$P = a+b+c = \frac{4}{3} - \frac{1}{6} + \frac{1}{6} = \frac{4}{3}$