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Evaluate:`int_2^3(2x^5+x^4-2x^3+2x^2+1)/((x^2+1)(x^4-1))dx` |
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Answer» Correct Answer - `(1)/(2)log6-(1)/(10)` Let `I= int_(2)^(3)(2x^(5)+x^(4)-2x^(3)+2x^(2)+1)/((x^(2)+1)(x^(4)-1))dx` `=int_(2)^(3)(2x^(5)-2x^(3)+x^(4)+1+2x^(2))/((x^(2)+1)(x^(2)-1)(x^(2)+1))dx` `=int_(2)^(3)(2x^(3)(x^(2)-1)+(x^(2)+1)^(2))/((x^(2)+1)^(2)(x^(2)-1))dx` `int_(2)^(3)(2x^(3)(x^(2)-1))/((x^(2)+1)^(2)(x^(2)-1))dx+int_(2)^(3)((x^(2)+1)^(2))/((x^(2)+1)^(2)(x^(2)-1))dx` `=int_(2)^(2)(2x^(3))/((x^(2)+1)^(2))dx+int_(2)^(3)(1)/((x^(2)-1))dx` `rArr I=I_(1)+I_(2)` Now , `I_(1)=int_(2)^(3)(2x^(3))/((x^(2)+1)^(2))dx` Put `x^(2)=1=trArr2x dx = dt` `:. I_(1)= int_(5)^(10)((t-1))/(t^(2))dt=int_(5)^(10)(1)/(t)dt-int(1)/(t^(2))dt` `=[logt _(5)^(10)+[(1)/(t)]_(5)^(10)` `=log10- log 5+(1)/(10)-(1)/(5)` `= log2 -(1)/(10)` Again , `I_(2)=int _(2)^(3)(1)/((x^(2)-1))dx =int_(2)^(3)(1)/((x-1)(x+1))dx` `=(1)/(2)int_(2)^(3)(1)/((x-1))dx -(1)/(2) int _(2)^(3)(1)/((x-1)(x+1))dx` `=[(1)/(2)log(x-1)]_(2)^(3)-(1)/(2)["log"(x+1)]_(2)^(3)` `=(1)/(2)" log " (2)/(1)" log " (4)/(3)` From Eq . (i) , `I=I_(1)+I_(2)` ` =" log " 2 - (1)/(10)+(1)/(2)" log " 2 -(1)/(2)" log " (4)/(3)` `=log [2*2^(1//2)((4)/(3))^(-1//2)]-(1)/(10)=(1)/(2) log 6 - (1)/(10)` |
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