InterviewSolution
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InterviewSolution
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Evaluate: `int((x^2+1)(x^2+2))/((x^2+3)(x^2+4)) dx` |
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Answer» We have `((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))=((t+1)(t+2))/((t+3)(t+4)),"where "x^(2)=t` `=((t^(2)+3t+2))/((t^(2)+7t+12))=1-((4t+10))/((t+3)(t+4)).` `Let ((4t+10))/((t+3)(t+4))+(A)/((t+3))+(B)/((t+4))` `implies (4t+10)-=A(t+4)+B(t+3).. . . (i)` Putting `t=-3`in(i) ,we get `A=-2` Putting `t=-4`in(i) ,we get `B=6`. `therefore ((4t+10))/((t+3)(t+4))=(-2)/((t+3))+(6)/((t+4)).` `thus ,((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4)=((t+1)(t+2))/((t+3)(t+4)),"where"x^(2)=t` `=((t^(2)+3t+2))/(t^(2)+7t+12))+1-((4t+10))/((t+3)(t+4))` `=1-{(-2)/((t+3))+(6)/((t+4))}["from(ii)"]` `={1+(2)/((t+3))-(6)/((t+4))}` `={1+(2)/((x^(2)+3))-(6)/((x^(2)+4))}.` `thereforeint((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx=int{1+(2)/((x^(2)+3))-(6)/((x^(2)+4))}dx.` `intdx+2int(dx)/((x^(2)+3))-6int(dx)/((x^(2)+4))` `=x+(2)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))-(6)/(2)tan^(-1)((x)/(2))+C` `=x+(2)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))-3tan^(-1)((x)/(2))+C.` |
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