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Evaluate:`int(x^2+4)/(x^4+16)dx` |
Answer» `I=int(x^(2)+4)/(x^(4)+16)dx` `=int(1+(4)/(x^(2)))/(x^(2)+(16)/(x^(2)))dx` `=int(1+(4)/(x^(2)))/(x^(2)+((4)/(x))^(2)-8+8)dx` `=int(1+(4)/x^(2))/((x-(4)/(x))^(2)+8)dx` Let `x-(4)/(x)=t. " Then " d(x-(4)/(x))=dt " or " (1+(4)/(x^(2)))dx=dt.` Therefore, `I=int(dt)/(t^(2)+(2sqrt(2))^(2))=(1)/(2sqrt(2))tan^(-1)((t)/(2sqrt(2)))+C` `=(1)/(2sqrt(2))tan^(-1)((x-(4)/(x))/(2sqrt(2)))+C` `=(1)/(2sqrt(2))tan^(-1)((x^(2)-4)/(2x sqrt(2)))+C` |
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