# evaluate the integral: ∫(9x^2-9x+9)/(x^2+1)dx

We can first factor out a 9 from the top.

So, we will have:

`int9((x^2 - x + 1)/(x^2+1))dx = 9int((x^2 - x + 1)/(x^2+1))dx`

We can apply Long Division to the integrand, we will have:

`9int(1 - x/(x^2 + 1))dx`

Then, we can write the integrand as a sum...

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We can first factor out a 9 from the top.

So, we will have:

`int9((x^2 - x + 1)/(x^2+1))dx = 9int((x^2 - x + 1)/(x^2+1))dx`

We can apply Long Division to the integrand, we will have:

`9int(1 - x/(x^2 + 1))dx`

Then, we can write the integrand as a sum of two integrands.

`9intdx - 9int(-x/(x^2+1))dx`

We will now proceed on the Integration.

For the first part we know that `intdu = u + c`

For the second part we will use U-substitution.

Let set `u = x^2 + 1`

So, `du = 2xdx`

Divide both sides of that by 2.

`(du)/2 = xdx`

We can now replace the u x^2 + 1 on the second term by u, and the xdx by du/2.

So, we will have:

`9x + c - 9int(du)/(2u) = 9x + c - 9/2int(du)/u`

So, we will have:

`9x + c - 9/2lnu + c`

Replace now the u by (x^2 + 1), and combine c + c = 2c.

`9x -9/2ln(x^2 + 1) + 2c`

Since, c is any constant, we can let C = 2c.

So, final answer here will be:

`9x - 9/2ln(x^2 + 1) + C`

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