Integrate the rational functions.

Let 
x = (Ax + B)(x -1) + C (x2 + 1)
⇒ x = Ax2 – Ax + Bx – B + Cx2 + C
Equating the coefficients of x2, x and constant term, we get,
A + C = 0
-A + B = 0
-B + c = 0
On solving these equation, we get,
A =
, B =
and C = ![]()
Thus,


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Now, let us consider,
,
Let (x2 + 1) = t
2xdx = dt
Thus,
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Therefore,

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