Giả sử a, b là hai số nguyên thỏa mãn \(\int\limits_1^5 {\dfrac{{dx}}{{x\sqrt {3x + 1} }} = a\ln 3 + b\ln 5} \). Tính giá trị của biểu thức \(P = {a^2} + ab + 3{b^2}.\)

\(\dfrac{{250}}{3}\;{\rm{m}}\).

\(270\;{\rm{m}}\).

\(200\;{\rm{m}}\).

\(\dfrac{{110}}{3}\;{\rm{m}}\).

\(S = \dfrac{{20}}{3}\)

\(S = \dfrac{{496}}{{15}}\)

\(S = \dfrac{4}{3}\)   

\(S = \dfrac{5}{3}\)

\(\int\limits_a^b {f\left( x \right)dx}  + \int\limits_b^c {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx} \)                        

\(\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}  = \int\limits_a^b {f\left( x \right)dx}  + \int\limits_a^b {g\left( x \right)dx} \)

\(\int\limits_a^b {f\left( {k.x} \right)dx}  = k\int\limits_a^b {f\left( x \right)dx} \)                         

\(\int\limits_a^b {k.f\left( x \right)dx}  = k.\int\limits_a^b {f\left( x \right)dx} \)

\(\int\limits_{}^{} {f\left( x \right)dx}  = \dfrac{{{2^x}}}{{\ln 2}} + \sin x + x + C\)

\(\int\limits_{}^{} {f\left( x \right)dx}  = \dfrac{{{2^x}}}{{\ln 2}} - \sin x + x + C\)

\(\int\limits_{}^{} {f\left( x \right)dx}  = {2^x}.\ln 2 + \sin x + x + C\)      

\(\int\limits_{}^{} {f\left( x \right)dx}  = {2^x}.\ln 2 - \sin x + x + C\)

\(S = 8\)

\(S = \dfrac{{220}}{3}\)

\(S = \dfrac{{76}}{3}\)

\(S = \dfrac{{148}}{3}\)

\(T = \dfrac{1}{2}\)

\(T = 0\)

\(T = 1\)

\(T = 2\)