InterviewSolution
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InterviewSolution
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`int1/(3x^2+13x-10)dx` |
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Answer» We have `(3x^(2)+13x-10)=3(x^(2)+(13)/(3)x-(10)/(3))` `=3{(x+(13)/(6))^(2)-(169)/(36)-(10)/(3)}=3{(x+(13)/(6))^(2)-(289)/(36)}` `=3{(x+(13)/(6))^(2)-((17)/(6))^(2)}`. `thereforeint(dx)/((3x^(2)+13x-10))=int(dx)/(3{(x+(13)/(6))^(2)-((17)/(6))^(2)})` `=(1)/(3)int(dt)/({t^(2)-((17)/(6))^(2)}) , "where" (x+(13)/(6))=t` `=(1)/(3)*(1)/((2xx(17)/(6)))log|(t-(17)/(6))/(t+(17)/(6))|+C[becauseint(dx)/((x^(2)-a^(2)))=(1)/(2a)log|(x-a)/(x+a)|]` `=(1)/(17)log|(6t-17)/(6t+17)|+C` `=(1)/(17)log|(6(x+(13)/(6))-17)/(6(x+(13)/(6))+17)|=(1)/(17)log|(6x-4)/(6x+30)|+C` `=(1)/(17){log(1)/(3)+log|(3x-2)/(x+5)|}+C` `=(1)/(17)log|(3x-2)/(x+5)|+k "where" (1)/(17)log(1)/(3)+C=k=` constant. |
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