∫x3x+1dx is equal to
x+x22+x33−ln|1+x|+C
x+x22−x33−ln|1+x|+C
x−x22−x33−ln|1+x|+C
x−x22+x33−ln|1+x|+C
x3x+1=x3+1−1x+1=−1(x+1)+x2+1−x
Thus given integral
=∫x2+1−x−11+xdx=x33+x−x22−ln|x+1|+C