Cho I = tích phân từ 0 đến 1 ( x + căn bậc 2 của x^2 + 15 ) d x = a + b ln 3 + c ln 5 với a , b , c thuộc Q . Tính tổng a+b+c.

\[I = \mathop \smallint \limits_0^1 \left( {x + \sqrt {{x^2} + 15} } \right)dx = \mathop \smallint \limits_0^1 xdx + \mathop \smallint \limits_0^1 \sqrt {{x^2} + 15} dx\]

\[{I_1} = \int\limits_0^1 {xdx} = \frac{1}{2}{x^2}\left| {_0^1} \right. = \frac{1}{2}\]

\({I_2} = \int\limits_0^1 {\sqrt {{x^2} + 15} } dx = x\sqrt {{x^2} + 15} \left| {_0^1} \right. - \int\limits_0^1 {x.\frac{x}{{\sqrt {{x^2} + 15} }}dx} \)

\[ = 4 - \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 15} }}dx = 4 - \int\limits_0^1 {\sqrt {{x^2} + 15} dx + \int\limits_0^1 {\frac{{15}}{{\sqrt {{x^2} + 15} }}dx} } } \]

\( \Rightarrow 2{I_2} = 4 + 15\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 15} }}} dx\)

Đặt

\[x + \sqrt {{x^2} + 15} = t \Rightarrow \left( {1 + \frac{x}{{\sqrt {{x^2} + 15} }}} \right)dx = dt \Leftrightarrow \frac{{dx}}{{\sqrt {{x^2} + 15} }} = \frac{{dt}}{t}\]

Khi đó:

\(\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 15} }}} dx = \int\limits_{\sqrt {15} }^5 {\frac{{dt}}{t}} = \ln \left| t \right|\left| {_{\sqrt {15} }^5} \right. = ln5 - ln\sqrt {15} = \frac{1}{2}\ln 5 - \frac{1}{2}\ln 3\)

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

\[I = {I_1} + {I_2} = \frac{1}{2} + 2 + \frac{{15}}{4}\ln 5 - \frac{{15}}{4}\ln 3 = \frac{5}{2} + \frac{{15}}{4}\ln 5 - \frac{{15}}{4}\ln 3 \Rightarrow a + b + c = \frac{5}{2} + \frac{{15}}{4} - \frac{{15}}{4} = \frac{5}{2}\]