Question

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

Answer 1

`int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x`

Explanation:

As in the given algebraic fraction, we have the only denominator `1+x^2`, which cannot be factorized. As it is a quadratic function, in partial decomposition, the numerator would be in form `ax+b`. But it is already in this form. Hence we cannot convert them into further partial fractions.

Now let `u=1+x^2`, hence `du=2xdx`

Hence, `int(x-1)/(1+x^2)dx`

= `intx/(1+x^2)dx-int1/(1+x^2)dx`

Now `intx/(1+x^2)dx=int(du)/(2u)=1/2xxlnu=(ln(1+x^2))/2`

and `int1/(1+x^2)dx=tan^(-1)x`

(as if `tan^(-1)x=v`, `x=tanv` and

`(dx)/(dv)=sec^2v=1+tan^2v=1+x^2`, hence `(dx)/(1+x^2)=dv`

and `int(dx)/(1+x^2)=intdv=v=tan^(-1)x`)

Hence `int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x`