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If `int(1)/((x+1)(x-2))dx=Alog_(e)(x+1)+Blog_(e)(x-2)+C` , then A + B = ?A. A+B=0B. A - B = 0C. AB =1D. AB =-1 |
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Answer» Correct Answer - a Let `I=int(1)/((x+1)(x-2))dx=int(1)/(3)((1)/(x-2)-(1)/(x+1))`dx `rArrI=(1)/(3)log_(e)(x-2)-(1)/(3)log_(e)(x+1)+C` `thereforeA=-(1)/(3)andB=(1)/(3)rArrA+B=0` |
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