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*Note that log here is the natural logarithm - by log(u) i mean ln(u).
We first perform long division on the integrand to get:
`int (1 + x/(1+x^2)) dx`
We then integrate the sum:
`int 1dx + int x/(1+x^2) dx`
`int dx + int x/(1+x^2) dx`
Then, we do substitution to solve the second integral. Let `u = 1+x^2` . This means that `du = 2xdx` and `xdx = (du)/2` . Substituting all these:
`int dx + int 1/(2u) du`
`int dx + (1/2) int (1/u) du`
The integrals are now easy to evaluate: `int dx = x +C_1` , while `1/2 int 1/u du = 1/2 log(u)+C_2.`
Therefore, we have:
`x + 1/2 log(u) + C` where C is the integration constant.
Plugging back `u = x^2 + 1` :
`int (x^2+x+1)/(x^2+1)dx= x + 1/2 log(x^2 + 1) + C`
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