Comments by "Fhf Fhf" (@fhffhff) on "Клинический случай"
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$dx/x²/√(1+x²)=$±(u²-1)^-0,5duu/(u²-1)/u=$±(u²-1)^-1,5du=-+(u²-1)^-0,5/u-$±(u²- 1)^-0,5u^-2du=-1/√(1+x²)/x-0,5x/√(1+x²) ^3-...-x^(2n-3)/n!/(1+x²)^(n-0,5)-...+C
√(1+x²)=u,x=±√(u²-1),dx=±0,5(u²-1)^-0,5*2udu
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