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∫5x+6/(x^2+x+1)

分母的凑微分d(x²+x+1)=(2x+1)dx 所以将分子变形为 5x+6 =5*[(2x+1)-1]/2+6 =(5/2)(2x+1)+7/2 而x²+x+1=(x+1/2)²+3/4 所以∫ (5x+6)/(x²+x+1) dx = ∫ [(5/2)(2x+1)+7/2]/(x²+x+1) dx = (5/2)∫ (2x+1)/(x²+x+1) dx ...

解: ∫1/(x²-5x+6)dx =∫[1/(x-3) -1/(x-2)]dx =ln|x-3|-ln|x-2|+C =ln|(x-3)/(x-2)| +C

明显的,后一张图片的答案打印错了, (x-2) 应为 (x+1);x 的取值范围是两个级数收敛域的交集。

x^2+5x+6= (x+2)(x+3) 1/(x^2+5x+6) = 1/(x+2) - 1/(x+3) y1 = 1/(x+2) y1' = -1/(x+2)^2 y1^(n) = (-1)^n. n!/(x+2)^(n+1) y2 = 1/(x+3) y2' = -1/(x+2)^2 y2^(n) = (-1)^n. n!/(x+3)^(n+1) y^(n) = y1^(n)+y2^(n) =(-1)^n. n! [ 1/(x+2)^(n+1)...

1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20) =1/x(x+1)+ 1/(x+1)(x+2) + 1/(x+2)(x+3)+ 1/(x+3)(x+4)+ 1/(x+4)(x+5) =[1/x - 1/(x+1)]+[1/(x+1) - 1/(x+2)]+[1/(x+2) - 1/(x+3)]+[1/(x+3) - 1/(x+4)]+[1/(x+4) - 1/...

原不等式化为[(x-½)²+¾](x+2)(x+3)(x-2)(x-2)>0. 第一项大于0,不考虑。讨论后四个。 同正 x>2 同付 x<-3 2正2负-2<x<2 综上x取值为x<-3或-2<x<2或x>2。

利用常见函数的幂级数展开 1/(1-x) = Σ[n=(0,∝)] x^n,x∈(-1,1) 所以f(x)=1/(x^2+5x+6) =1/[(x+2)(x+3)] =1/(x+2) - 1/(x+3) =1/[6+(x-4)] - 1/[7+(x-4)] =(1/6) * 1/[1+(x-4)/6] - (1/7) * 1/[1+(x-4)/7] =(1/6) * 1/[1-(-1)*(x-4)/6] - (1/7) ...

x^2+x-2/x^2+5x+6 =(x+2)(x-1)/(x+2)(x+3) =(x-1)/(x+3) =(-2-1)/(-2+3) =-3

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