# 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?

I got this integral:

Now at the **step 2** we take out

But why we don't do the same thing at the **step 4**?

So:

and so on?

##### 2 Answers

Because the denominator would be

#### Explanation:

In order to use

Attempting to eliminate the factor

So instead, we use

There is a small arithmetic error here:

The correct working would be:

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

So to answer you question, to factor that

*- A small recommendation:*

There is a standard integral that:

Given the substitution:

you could use this to evaluate the integral from directly from step 1.