Solutions - 0060-05

(1) Numerator degree is not smaller! Long division first:

$$\frac{x^2-x+1}{x^2+x}\gg \gg 1+\frac{-2x+1}{x^2+x}$$

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(2) Factor denominator:

$$x^2+x\gg \gg x(x+1)$$

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(3) Write the partial fractions general form equation (for the second term):

$$\frac{-2x+1}{x^2+x}=\frac{A}{x}+\frac{B}{x+1}$$

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(4) Solve for constants:

Cross multiply:

$$\gg \gg -2x+1=A(x+1)+Bx$$

Plug in $x=0$, obtain $1=A\cdot 1$ so $A=1$.

Plug in $x=-1$, obtain $3=B(-1)$ so $B=-3$.

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(5) Integrate by terms:

$$\begin{array}{cc}\int 1+\frac{-2x+1}{x^2+x}dx& \gg \gg \int dx+\int \frac{1}{x}dx+\int \frac{-3}{x+1}dx\\ & \\ & \gg \gg x+\ln |x|-3\ln |x+1|+C\end{array}$$