Cho tích phân 1 đến 3 dx/(x+1)(x+4)=aln2+bln5+cln7. Tính S=a+4b-c

Ta có $\frac{1}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = \frac{1}{3}\left[ {\frac{1}{{x + 1}} - \frac{1}{{x + 4}}} \right]$.

Do đó:

$\begin{array}{l}\int_1^3 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}}  = \frac{1}{3}\ln \left| {\frac{{x + 1}}{{x + 4}}} \right|\left| \begin{array}{l}3\\1\end{array} \right. = \frac{1}{3}\left( {\ln \frac{4}{7} - \ln \frac{2}{5}} \right)\\ = \frac{1}{3}\left( {\ln 2 + \ln 5 - \ln 7} \right) = \frac{1}{3}\ln 2 + \frac{1}{3}\ln 5 - \frac{1}{3}\ln 7 \Rightarrow \left\{ \begin{array}{l}a = \frac{1}{3}\\b = \frac{1}{3}\\c =  - \frac{1}{3}\end{array} \right.\end{array}$

Vậy: $S = \frac{1}{3} + 4.\frac{1}{3} - \frac{1}{3} = \frac{4}{3}$