Biết \(I = \int\limits_3^4 {\frac{{{\rm{d}}x}}{{{x...

B

- Hướng dẫn giải

Cách 1:

\(I = \int\limits_3^4 {\frac{{{\rm{d}}x}}{{{x^2} + x}}} \). Ta có: \(\frac{1}{{{x^2} + x}} = \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{x} - \frac{1}{{x + 1}}.\)

Khi đó: \(I = \int\limits_3^4 {\frac{{{\rm{d}}x}}{{{x^2} + x}}}  = \int\limits_3^4 {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right){\rm{d}}x = } \left. {\left[ {\ln x - \ln \left( {x + 1} \right)} \right]} \right|_3^4 = \left( {\ln 4 - \ln 5} \right) - \left( {\ln 3 - \ln 4} \right) = 4\ln 2 - \ln 3 - \ln 5.\)

Suy ra: \(a = 4,b =  - 1,c =  - 1.\) Vậy S = 2

Cách 2: Casio

Ta có: \(I = \int\limits_3^4 {\frac{{{\rm{d}}x}}{{{x^2} + x}} = a\ln 2 + b\ln 3 + c\ln 5}  \Rightarrow {{\rm{e}}^I} = {{\rm{e}}^{a\ln 2 + b\ln 3 + c\ln 5}} = {{\rm{e}}^{\ln \left( {{2^a}{{.3}^b}{{.5}^c}} \right)}}\) \( \Rightarrow {{\rm{e}}^I} = {2^a}{.3^b}{.5^c}\) 

Hay \(\frac{{16}}{{15}} = {2^a}{.3^b}{.5^c} \Leftrightarrow {2^4} = {2^a}{.3^{b + 1}}{.5^{c + 1}} \Rightarrow \left\{ \begin{array}{l}
a = 4\\
b + 1 = 0\\
c + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 4\\
b =  - 1\\
c =  - 1
\end{array} \right. \Rightarrow S = a + b + c = 2\).