Question

How do i integrate `int (1-x^2)/(4+x^2) dx`?

Answer 1

Given: `int (1-x^2)/(4+x^2) dx`

Factor out -1 from the numerator:

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

Add zero to the numerator in the form of `+4-4`:

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

Combine the negatives:

`int (1-x^2)/(4+x^2) dx = -int (x^2+4-5)/(x^2+4) dx`

Separate into two fractions:

`int (1-x^2)/(4+x^2) dx = -int (x^2+4)/(x^2+4)-5/(x^2+4) dx`

The first fraction becomes 1:

`int (1-x^2)/(4+x^2) dx = -int 1-5/(x^2+4) dx`

Separate into two integrals:

`int (1-x^2)/(4+x^2) dx = int 5/(x^2+4) dx -intdx`

Bring the 5 outside the integral:

`int (1-x^2)/(4+x^2) dx = 5int 1/(x^2+4) dx -intdx`

Both integrals are well known:

`int (1-x^2)/(4+x^2) dx = 5/2tan^-1(x/2)-x+C`