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Answer 1 · 2017-05-19T21:02:51

#int (3x+5)/(x^3-x^2-x+1)dx=1/2lnabs((x+1)/(x-1))-4/(x-1)+"c"#

#int (3x+5)/(x^3-x^2-x+1)# #dx#

#=int(3x+5)/((x-1)^2(x+1))# #dx#

Perform partial fraction decomposition on the integrand

#(3x+5)/((x-1)^2(x+1))-=A/(x-1)^2+B/(x-1)+C/(x+1)#

#3x+5-=A(x+1)+B(x-1)(x+1)+C(x-1)^2#

Let #x=1#

#8=2ArArrA=4#

Let #x=-1#

#2=4CrArrC=1/2#

Let #x=0#

#5=A-B+C#

#B=-1/2#

#(3x+5)/((x-1)^2(x+1))=4/(x-1)^2-1/(2(x-1))+1/(2(x+1))#

The integral is now:

#int(4/(x-1)^2-1/(2(x-1))+1/(2(x+1)))# #dx#

#=1/2lnabs(x+1)-1/2ln|x-1|-4/(x-1)+"c"#

#=1/2lnabs((x+1)/(x-1))-4/(x-1)+"c"#