Question

I don't get my mistake on how to solve `2int (1)/(x^2-x+1)dx`, can you help me?

I got this integral:

Now at the step 2 we take out `3/4`, invert it in `4/3` and multiply with `2`.

But why we don't do the same thing at the step 4?
So:
`8/3int1/(1/3(2x-1)^2+1)dx`
`3*8/3int1/((2x-1)^2+1)dx`
`8int1/((2x-1)^2+1)dx`
and so on?

Answer 1

Because the denominator would be `(2x-1)^2+3` and not `(2x-1)^2+1`

Explanation:

In order to use `int 1/(t^2+1) dt = tan^(-1)t+C` we need the denominator to be of the form `t^2+1`.

Attempting to eliminate the factor `1/3` from `1/3(2x-1)^2+1` by multiplying by `3` would result in `(2x-1)^2+3`, which is not what we want.

So instead, we use `1/3(2x-1)^2 = ((2x-1)/sqrt(3))^2`

Answer 2

There is a small arithmetic error here:

`8/3int1/(1/3(2x-1)^2+1)dxcolor(red)ne3*8/3int1/((2x-1)^2+1)dx`

The correct working would be:

`8/3int1/(1/3(2x-1)^2+1)dx=3*8/3int1/((2x-1)^2color(red)(+3))dx`

A similar error is also made in going from step 3 to step 4.

So to answer you question, to factor that `1/3` from the bottom would require multiplying that `1` by 3, meaning that the direct integration would no longer work.

- A small recommendation:

There is a standard integral that:

`int1/(u^2+a^2)du=1/aarctan(u/a)+C`

Given the substitution:
`u=x-1/2`
you could use this to evaluate the integral from directly from step 1.