Solutions - 0060-07

(1) Denominator has degree 3, numerator has degree 2, therefore long division is not necessary.

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(2) Write the partial fractions general form equation:

Notice that $x^2+4$ is an irreducible quadratic (cannot be factored). So we have:

$$\frac{5x^2-5x+14}{(x-2)(x^2+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+4}$$

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(3) Solve for constants:

Cross multiply:

$$\gg \gg 5x^2-5x+14=A(x^2+4)+(Bx+C)(x-2)$$

Plug in $x=2$, obtain:

$$20-10+14=A(8)\gg \gg 24=8A\gg \gg A=3$$

Expand RHS:

$$\begin{array}{cc}5x^2-5x+14& =3x^2+12+B{x}^2-2Bx+Cx-2C\\ & \\ & =(3+B)x^2+(-2B+C)x+(12-2C)\end{array}$$

Comparing $x^2$ terms, obtain: $5=3+B$ and thus $B=2$.

Comparing constant terms, $C=-1$.

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(4) Integrate by terms:

$$\begin{array}{ccc}& \int \frac{5x^2-5x+14}{(x-2)(x^2+4)}dx\gg \gg \int \frac{3}{x-2}dx+\int \frac{2x-1}{x^2+4}dx& \\ & & \\ & \gg \gg 3\ln |x-2|+\int \frac{2x}{x^2+4}dx-\int \frac{1}{x^2+4}dx& \\ & & \\ & \gg \gg 3\ln |x-2|+\ln |x^2+4|-\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C& \text{(Note A)}\end{array}$$

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Note A: For the last term, use the formula:

$$\int \frac{dx}{x^2+h^2}=\frac{1}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$