Evaluate`int (dx)/((x^(3)+x^(2)+x+1)).`

1.	Evaluate`int (dx)/((x^(3)+x^(2)+x+1)).`
Answer» We have `(1)/((x^(3)+x^(2)+x+1))=(1)/(x^(2)(x+1)+(x+1))=(1)/((x+1)(x^(2)+1)).` `Let (1)/((x+1)(x^(2)+1))=(A)/((x+1))+(Bx+C))/((x^(2)+1))` `implies 1-=A(x^(2)+1)+(Bx+C)(x+1).. . . .(i)` Putting `x=-1`on both sides of (i),we get `A=(1)/(2).` Comparing the coefficients of `x^(2)` on both sides of (i) , we get `A+B=0impliesB=-A=(-1)/(2).` COmparing the coefficients of x on both sides of (i) , we get `B+C=0impliesC=-B=(1)/(2).` `therefore (1)/((x+1)(x^(2)+1))=(1)/(2(x+1))+((-1)/(2)x+(1)/(2))/(x^(2)+1)` `therefore int(dx)/((x^(3)+x^(2)+x+1))=int(dx)/((x+1)(x^(2)+1))` `=(1)/(2).int(dx)/((x+1))-(1)/(2)int(x)/((x^(2)+1))dx+(1)/(2)(dx)/((x^(2)+1))` `=(1)/(2).int(dx)/((x+1))-(1)/(4).int(2x)/((x^(2)+1))dx+(1)/(2)int(dx)/((x^(2)+1))` `=(1)/(2)log \|x+1\|-(1)/(4) log \|x^(2)+1\|+(1)/(2) tan^(-1)x+C.`

Discussion

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