intgeral of : 72x/72x-7x+1 dx Share with your friends Share 0 S. Usha Rani answered this ∫72x72x-7x+1dxPut 7x=tThen 7xlog7dx=dtt log7dx=dtdx=dttlog7Then ∫t2t2-t+1dttlog7=1log7∫tt2-t+1dt=12log7∫2tt2-t+1dt(multiplying and dividing by2)=12∫2t-1+1t2-t+1dt(adding and subtracting 1)=12log7∫2t-1t2-t+1dt+12log7∫1t2-t+1dt ( we have t2-t+1= t2-2*t*12+14-14+1= (t-12)2+(32)2)=12log7log(t2-t+1)+12log7∫1(t-12)2+(32)2=12log7log(t2-t+1)+12log7132tan-1t-1232=12log7log(t2-t+1)+13log7tan-12t-13=1log 49log(72x-7x+1)+13log7tan-12*7x-13 0 View Full Answer