Biết tích phân 1 đến 2 xdx/(x+1)(2x-1)= aln 2 + bln 3 + cln 5}. Tính S=a+b+c

Ta có:

\(\begin{array}{l} \int\limits_1^2 {\frac{{x{\rm{d}}x}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}}} = \int\limits_1^2 {\left[ {\frac{{\rm{1}}}{{\left( {x + 1} \right)}} - \frac{1}{{\left( {2x + 1} \right)}}} \right]{\rm{d}}x} \\ = \left. {\left( {\ln \left| {x + 1} \right| - \frac{1}{2}\ln \left| {2x + 1} \right|} \right)} \right|_1^2\\ = \ln 3 - \frac{1}{2}\ln 5 - \ln 2 + \frac{1}{2}\ln 3 = - \ln 2 + \frac{3}{2}\ln 3 - \frac{1}{2}\ln 5. \end{array}\)

Vậy S=0.