How do you integrate #int (x-1)/ (x^3 +x)# using partial fractions?

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Answer 1 · 2017-01-05T08:24:12

The answer is #=-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C#

Let's factorise the denominator

#x^3+x=x(x^2+1)#

So,

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

#=A/x+(Bx+C)/(x^2+1)#

#=(A(x^2+1)+((Bx+C)x))/(x(x^2+1))#

Therefore,

#x-1=A(x^2+1)+x(Bx+C)#

Let #x=0#, #=>#, -1=A#

Coefficients of #x^2#

#0=A+B#, #=>#, #B=1#

Coefficients of #x#

#1=C#

So,

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

Therefore,

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

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

We calculate each integral separately

#-intdx/x=-ln(∣x∣)#

Let #u=x^2+1#, #=>#, #du=2xdx#

#int(xdx)/(x^2+1)=int(2du)/u=1/2lnu#

#=1/2ln(x^2+1)#

Let #x=tantheta#, #=>#, #dx=sec^2theta d theta#

and #x^2+1=tan^2theta + 1=sec^2theta#

So,

#int(1dx)/(x^2+1)=int(sec^2theta d theta)/sec^2theta=intd theta=theta#

#=arctanx#

Putting it alltogether,

#int((x-1)dx)/(x^3+x)=-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C#