Cho \(I = \int\limits_0^1 {\left( {x + \sqrt {{x^2} + 15} } \right)dx}  = a + b\ln 3 + c\ln 5\) với \(a,b,c \in \mathbb{Q}\). Tính tổng \(a + b + c\).

\(I = \int\limits_0^1 {\left( {x + \sqrt {{x^2} + 15} } \right)dx}  = \int\limits_0^1 {xdx}  + \int\limits_0^1 {\sqrt {{x^2} + 15} dx} \)

\({I_1} = \int\limits_0^1 {xdx}  = \left. {\dfrac{1}{2}{x^2}} \right|_0^1 = \dfrac{1}{2}\)

\(\begin{array}{l}{I_2} = \int\limits_0^1 {\sqrt {{x^2} + 15} dx}  = \left. {x\sqrt {{x^2} + 15} } \right|_0^1 - \int\limits_0^1 {x.\dfrac{x}{{\sqrt {{x^2} + 15} }}dx} \\\,\,\,\,\,\, = 4 - \int\limits_0^1 {\dfrac{{{x^2}}}{{\sqrt {{x^2} + 15} }}dx}  = 4 - \int\limits_0^1 {\sqrt {{x^2} + 15} dx}  + \int\limits_0^1 {\dfrac{{15}}{{\sqrt {{x^2} + 15} }}dx} \\ \Rightarrow 2{I_2} = 4 + 15\int\limits_0^1 {\dfrac{1}{{\sqrt {{x^2} + 15} }}dx} \end{array}\)

Đặt \(x + \sqrt {{x^2} + 15}  = t \Rightarrow \left( {1 + \dfrac{x}{{\sqrt {{x^2} + 15} }}} \right)dx = dt \Leftrightarrow \dfrac{{dx}}{{\sqrt {{x^2} + 15} }} = \dfrac{{dt}}{t}\).

Khi đó: \(\int\limits_0^1 {\dfrac{1}{{\sqrt {{x^2} + 15} }}dx}  = \int\limits_{\sqrt {15} }^5 {\dfrac{{dt}}{t}}  = \left. {\ln \left| t \right|} \right|_{\sqrt {15} }^5 = \ln 5 - \ln \sqrt {15}  = \dfrac{1}{2}\ln 5 - \dfrac{1}{2}\ln 3\)

\( \Rightarrow 2{I_2} = 4 + 15.\left( {\dfrac{1}{2}\ln 5 - \dfrac{1}{2}\ln 3} \right) \Leftrightarrow {I_2} = 2 + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3\)

\(I = {I_1} + {I_2} = \dfrac{1}{2} + 2 + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3 = \dfrac{5}{2} + \dfrac{{15}}{4}\ln 5 - \dfrac{{15}}{4}\ln 3 \Rightarrow a + b + c = \dfrac{5}{2} + \dfrac{{15}}{4} - \dfrac{{15}}{4} = \dfrac{5}{2}\)