f the graph of the antiderivative F(x) of f(x)=log(logx)+(logx)−2 passes through (e,1998−e) then the term independent of x in F(x) is
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Solution
Antiderivative of f(x)=F(x)=∫(log(logx)+(logx)−2)dx+C =xlog(logx)−∫xxlogxdx+∫(logx)−2dx+C=xlog(logx)−[x(logx)−1+∫(logx)−2dx]+∫(logx)−2dx+C=xlog(logx)−x(logx)−1+C For point (e,1998−e), we get 1998−e=e.0−e+C⇒C=1998