We find that, the #Dr.=x^3+x=x(x^2+1)#
This, the poly. of the #Dr.# has a linear factor #x#, and a non-reducible quadr. factor #x^2+1#. Therefore, we will suppose that
#(x-1)/(x^3+x)=(x-1)/(x(x^2+1))=A/x+(Bx+c)/(x^2+1), A,B,C in RR...(star)#.
But, upon simplification,
#The R.H.S.=(A(x^2+1)+(Bx+C)x)/(x(x^2+1))#.
Thus, #(x-1)/(x^3+x)=(A(x^2+1)+(Bx+C)x)/(x(x^2+1))... ......(1)#.
As the #Drs.# of #(1)# are same, so must be the #Nrs.#. Hence,
#Ax^2+A+Bx^2+Cx=(A+B)x^2+Cx+A=x-1......(2)#.
Comparing the resp. co-effs. of both sides, we immediately get,
#A=-1, C=1, and, A+B=0rArrB=-A=1#.
Therefore, by #(star)#,
#int(x-1)/(x^3+x)dx=int[-1/x+(x+1)/(x^2+1)]dx#
#=-int1/xdx+int(x+1)/(x^2+1)dx=-ln|x|+int{x/(x^2+1)+1/(x^2+1)}dx#
#=-ln|x|+1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx#
#=-ln|x|+1/2int(d/dx(x^2+1))/(x^2+1)dx+arctanx#
#=-ln|x|+1/2ln(x^2+1)+arctanx+C#, OR,
#=ln(sqrt(x^2+1)/|x|)+arctanx+C#.
In the final step, this well-known Rule has been used :-
The Rule# := int(f'(x))/f(x)dx=ln|f(x)+K#.
I hope, this will be of Help! Enjoy Maths.!
