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 $P = {a^2} + ab + 3{b^2}.$

$\dfrac{{250}}{3}\;{\rm{m}}$.

$270\;{\rm{m}}$.

$200\;{\rm{m}}$.

$\dfrac{{110}}{3}\;{\rm{m}}$.

$S = \dfrac{{20}}{3}$

$S = \dfrac{{496}}{{15}}$

$S = \dfrac{4}{3}$   

$S = \dfrac{5}{3}$

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

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

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

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

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

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

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

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

$S = 8$

$S = \dfrac{{220}}{3}$

$S = \dfrac{{76}}{3}$

$S = \dfrac{{148}}{3}$

$T = \dfrac{1}{2}$

$T = 0$

$T = 1$

$T = 2$