Quiz #6 Solution

Expand

$$\frac{3x^2 -2x +3}{(x^2-1)(x^2+1)}dx$$

As a partial fraction. Then evaluate

$$I=\int{ \frac{3x^2 -2x +3}{(x^2-1)(x^2+1)}dx}$$

Solution: Set

$$\frac{3x^2 -2x +3}{(x^2-1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$$

The heavy side method quickly gives A=1 and B=-2. After getting a common denominator and adding on the RHS:

$$\frac{3x^2 -2x +3}{(x^2-1)(x^2+1)} = \frac{(x+1)(x^2+1) -2(x-1)(x^2+1) + (Cx+D)(x-1)(x+1)}{(x^2-1)(x^2+1)}$$

Multiplying out the numerator on the RHS and setting equal to the numerator on the LHS gives:

3x²-2x+3 = x³(C-1) + x²(D+3) + x(-C-1) + (3-D)

From which we get C=1 and D=0. So the partial fraction decompostion is:

$$\frac{3x^2-2x+3}{(x^2-1)(x^2+1)} = \frac{1}{x-1} - \frac{2}{x+1} + \frac{x}{x^2+1}$$

Integrating the first two terms is easy:

$$\int{ \frac{dx}{x-1}} = ln\left\vert x-1\right\vert + c_1$$

$$\int{ \frac{-2dx}{x+1}} = -2ln\left\vert x+1\right\vert + c_2$$

And the substitution u=x²+1 gives the last term:

$$\int{ \frac{xdx}{x^2+1}} = \frac{1}{2}ln(x^2+1) + c_3$$

Thus:

$$I = ln\left\vert \frac{(x-1) \sqrt{x^2+1}}{(x+1)^2}\right\vert + c$$

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