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:
enter image source here

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?

2 Answers
George C.
Jan 15, 2018

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#

Andrew I.
Jan 15, 2018

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.

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