Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 74886 by abdomathmax last updated on 03/Dec/19

$$calculate\int \frac{x+1}{{(x^3+x-2)}^2}dx$$

Commented by mathmax by abdo last updated on 15/Dec/19

$$letI=\int \frac{x+1}{{(x^3+x-2)}^2}dxletdecomposeF\left(x\right)=\frac{x+1}{{(x^3+x-2)}^2}$$$$wehave{x}^3+x-2=x^3-1+x-1=(x-1)(x^2+x+1)+x-1$$$$=(x-1)(x^2+x+2)\Rightarrow F\left(x\right)=\frac{x+1}{{(x-1)}^2{(x^2+x+2)}^2}$$$$=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{cx+d}{x^2+x+2}+\frac{ex+f}{{(x^2+x+2)}^2}$$$$b={(x-1)}^{2}F\left(x\right){\mid }_{x=1}=\frac{2}{4^2}=\frac{1}{8}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{8{(x-1)}^2}+\frac{-ax+d}{x^2+x+2}+\frac{ex+f}{{(x^2+x+2)}^2}$$$$F\left(0\right)=\frac{1}{4}=-a+\frac{1}{8}+\frac{d}{2}+\frac{f}{4}\Rightarrow$$$$1=-4a+\frac{1}{2}+2d+f\Rightarrow -4a+2d+f=\frac{1}{2}$$$$F\left(2\right)=\frac{3}{64}=a+\frac{1}{8}+\frac{-2a+d}{8}+\frac{2e+f}{64}\Rightarrow$$$$3=64a+8+8(-2a+d)+2e+f\Rightarrow 3=48a+8+8d+2e+f\Rightarrow$$$$\Rightarrow 48a+8d+2e+f=-5$$$$weformasystemwith4equationwithunknowna,d,eandf$$$$togetthevalueanywayweget$$$$I=aln\mid x-1\mid -\frac{1}{8(x-1)}+\int \frac{-ax+d}{x^2)x+2}dx+\int \frac{ex+f}{{(x^2+x+2)}^2}$$$$...becontinued...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com