Question

How do you integrate `(x-1)/ (x^3 +x^2)` using partial fractions?

Answer 1

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.!