Question

Question #28fed

Answer 1

`-1/2ln(2x^2-2x+5)+C`

Explanation:

You wish to find:

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

Let `u=2x^2-2x+5`. This implies that `du=(4x-2)dx`.

Multiply the numerator of the integrand by `-2`. Balance this by multiplying the exterior of the integral by `-1//2`.

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

Substitute in `u=2x^2-2x+5` and `du=(4x-2)dx` into the integrand:

`=-1/2int(du)/u`

Note that `int(du)/u=lnabsu+C`:

`=-1/2ln(2x^2-2x+5)+C`

Note that the absolute value signs are not needed because `2x^2-2x+5>0` for all values of `x`.