InterviewSolution
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InterviewSolution
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`int(dx)/((1-6x-9x^2)` |
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Answer» We have `int(dx)/((1-6x-9x^(2)))=-int(dx)/((9x^(2)+6x-1))=-(1)/(9)int(dx)/((x^(2)+(2)/(3)x-(1)/(9)))` `=-(1)/(9)*int(dx)/({(x^(2)+(2)/(3)x+(1)/(9))-(2)/(9)})` `=-(1)/(9)*int(dx)/({(x+(1)/(3))^(2)-((sqrt(2))/(3))^(2)})=(1)/(9)*int(dx)/({((sqrt(2))/(3))^(2)-(x+(1)/(3))^(2)})` `=(1)/(9)*int(dx)/({((sqrt(2))/(3))^(2)-t^(2)}),"where"(x+(1)/(3))=t` `=(1)/(9)*(1)/(2*(sqrt(2))/(3))log|(sqrt(2)/(3)+t)/(sqrt(2)/(3)-t)|+C=(1)/(6sqrt(2))log|(sqrt(2)+3t)/(sqrt(2)-3t)|+C` `=(1)/(6sqrt(2))log|(sqrt(2)+3(x+(1)/(3)))/(sqrt(2)-3(x+(1)/(3)))|+C` `=(1)/(6sqrt(2))log|(sqrt(2)+1+3x)/(sqrt(2)-1-3x)|+C`. |
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