If f(x)=∫7x+x7−17xlog7+7x6dx then f(1)−f(0)
3
3 log 2
2 log 3
2
f(x)=∫7xlog7+7x6x7+7xdx
Let 7x+x7=t⇒7xlog7+7x6dx=dtf(x)=∫ 1tdt=logt+cf(x)=logx7+7x+c
f(1)−f(0)=log(8)−0=3log2