`int_0^1 x [ ln ( x + 2) - 1/( x ^2 + 1)] dx `
`= int_0^1 [x ln ( x + 2) - ( x )/( x ^2 + 1)] dx`
`= \underbrace{int_0^1 x ln ( x + 2) dx}_{I_1} - \underbrace{ int_0^1( x )/( x ^2 + 1) dx}_{I_2}`
Đặt `{(u=ln(x+2)),(dv=xdx):}=>{(du=1/(x+2)dx),(v=x^2/2dx):}`
Khi đó, $I_1=\mathop {\left. {\frac{{{x^2}}}{2}\ln (x + 2)} \right|}\nolimits_0^1 - \frac{1}{2}\int_0^1 {\frac{{{x^2}}}{{x + 2}}} dx$
$= \frac{{\ln 3}}{2} - \frac{1}{2}\int_0^1 {(x + \frac{4}{{x + 2}} - 2)} dx$
$ = \frac{{\ln 3}}{2} - \frac{1}{2}\int_0^1 {xdx - 2\int_0^1 {\frac{1}{{x + 2}}dx + \int_0^1 {1dx} } } $
$ = \frac{{\ln 3}}{2} - \left. {\frac{{{x^2}}}{4}} \right|_0^1 - \left. {2\ln \left| {x + 2} \right.} \right|_0^1 + \left. x \right|_0^1$
$=\frac{{\ln 3}}{2} - \frac{1}{4} - \ln \frac{9}{4} + 1$
Đặt `t=x^2+1=>dt=2xdx=>dx=(dt)/(2x),` thay vào `I_2`, ta được:
`-1/2int_1^2 1/tdt`
$=\left. { - \frac{{\ln \left| t \right|}}{2}} \right|_1^2$
`=-ln2/2`
Ta có: `I=I_1+I_2=\frac{\ln 3}{2} - \frac{1}{4} - \ln \frac{9}{4} + 1-\frac{\ln 2}{2}= 3/4 - 3/2 ln\frac{3}{2}`
$=3-6\ln\frac{3}{2}=\frac{6\ln2-6\ln3+3}{4}$
Mà `I=(abln2+bcln3-c)/4`
`<=> \frac{6\ln2-6\ln3+3}{4}=(abln2+bcln3-c)/4`
`<=> 6ln2-6ln3+3=abln2+bcln3-c`
`<=> {(ab=6),(bc=-6),(c=-3):}<=>{(a=3),(b=2),(c=-3):}` `(`loại vì `c notin NN)`
`=>` Không thể tìm bộ ba `a; b; c in NN`
`=>` Không tính được `A=a+b+c` với `a; b; c in NN`
