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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 12: Cho $I=\int_{0}^{1} x\left(\ln (x+2)-\frac{1}{x^{2}+1}\right) \mathrm{d} x=\frac{a b \ln 2+b c \ln 3-c}{4}$ với $a,b,c\in N$ Tính $A=a

Toán Lớp 12: Cho $I=\int_{0}^{1} x\left(\ln (x+2)-\frac{1}{x^{2}+1}\right) \mathrm{d} x=\frac{a b \ln 2+b c \ln 3-c}{4}$ với $a,b,c\in N$ Tính $A=a+b+c$

Comments ( 1 )

  1. 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

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222-9+11+12:2*14+14 = ? ( )

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