∫10x log(1+2x)dx
Let I=∫10xlog(1+2x)dx
=[log(1+2x)x22]10−∫11+2x.2.x22dx=12[x2log(1+2x)]10−∫x21+2xdx=22[1log 3−0]−[∫10(x2−x21+2x)dx]=12log 3−12∫10x dx+12∫10x1+2xdx=12log 3−12[x22]10+12∫1012(2x+1−1)(2x+1)dx=12log 3−12[12−0]+14∫10dx−14∫1011+2xdx=12−14+14[x]10−18[log|(1+2x)|]10=12−14+14−18[log3−log1]=22−18log 3=1−18log 3