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int[1/(Inx)-1/(Inx)^2]dx |
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Answer» Solution :`int[1/(Inx)-1/(Inx)^2]dx` =`int1/(Inx).1-INTDX/(Inx)^2` [INTEGRATING by parts taking `1/(Inx)` as first function and 1 as second function.] =`1/(Inx).x-int(-1)/(Inx)^2 . 1/x.xdx-intdx/(Inx)^2+C` =`x/(Inx)+intdx/(Inx)^2-intdx/(Inx)^2+C` `x/(Inx)+C` |
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