1.

int[1/(Inx)-1/(Inx)^2]dx

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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