First, I check if the rational function is proper. If it isn’t, I need to do polynomial long division. This is already proper, so no long division is necessary.
Next, I write the partial fraction decomposition with unknowns. I take the right side to common denominator and group the coefficients of the numerator.
\begin{align*}
\amp \frac{2x^2+4x-1}{(x^2+3x+4)(x^2+6)} = \frac{ax +
b}{x^2+3x+4} + \frac{cx+d}{x^2+6} \\
\amp = \frac{(ax+b)(x^2+6) +
(cx+d)(x^2+3x+4)}{(x^2+3x+4)(x^2+6)}\\
\amp = \frac{ax^3 + bx^2 + 6ax + 6b + cx^3 + dx^2 + 3cx^2 +
3dx + 4cx + 4d}{(x^2+3x+4)(x^2+6)} \\
\amp = \frac{(a+c)x^3 + (b + d + 3c)x^2 + (6a + 4c + 3d) x +
(6b + 4d)}{(x^2+3x+4)(x^2+6)}
\end{align*}
I compare the numerator coefficients to get the system of equations. I proceed to solve the system. (I haven’t anotated all the steps of isolating and replacing variables.)
\begin{align*}
a+c \amp = 0 \\
b +d + 3c \amp = 2 \\
6a + 4c + 3d \amp = 4 \\
6b + 4d \amp = -1
\end{align*}
For these late questions, I’m leaving out the details of solving the system. For this one, I asked a computer to solve the system for me. Here is the solution.
\begin{align*}
a \amp = \frac{-31}{58}
\amp \amp b = \frac{-75}{58}
\amp \amp c = \frac{31}{58}
\amp \amp d = \frac{98}{58}
\end{align*}
The numerator coefficients give me the partial fraction decomposition. Now that I have completed partial fractions, I integrate the pieces using the four basic rational function integrals. I’ll do the two pieces of the integral separately and then combine them. Here is the first piece.
\begin{align*}
\amp \frac{x^2+4x-1}{(x^2+3x+4)(x^2+6)} \frac{1}{58}
\left[ \frac{-31x - 75}{x^2+3x+4} + \frac{31x+98}{x^2+6}
\right] \\
\int \frac{-31x - 75}{x^2+3x+4} dx \amp = \frac{-31}{2} \int
\frac{ 2x + \frac{150}{31}}{x^2+3x+4} dx \\
\amp = \frac{-31}{2} \int \frac{2x + 3 + \frac{57}{31}}{x^2
+3 x + 4} dx\\
\amp = \frac{-31}{2} \int \frac{2x+ 3}{x^2 +3 x +4} dx +
\frac{-31}{2} \frac{57}{31} \int \frac{1}{x^2+3x+4}dx \\
\amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2} \int
\frac{1}{x^2 + 3x + \frac{9}{4} + \frac{7}{4}} dx \\
\amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2} \int
\frac{1}{ \left(x+\frac{3}{2} \right)^2 + \frac{7}{4}} dx \\
\amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2}
\frac{2}{\sqrt{7}} \arctan \left( \frac{x +
\frac{3}{2}}{\frac{\sqrt{7}}{2}} \right) + c \\
\amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{\sqrt{7}}
\arctan \left( \frac{2x + 3}{\sqrt{7}} \right) + c
\end{align*}
Here is the second piece.
\begin{align*}
\int \frac{31x+98}{x^2+6} dx \amp = \frac{31}{2} \int
\frac{2x + \frac{196}{31}}{x^2+6} dx \\
\amp = \frac{31}{2} \int \frac{2x}{x^2+6} dx + \frac{31}{2}
\frac{196}{31} \int \frac{1}{x^2 + 6} dx \\
\amp = \frac{31}{2} \ln |x^2+6| + 98 \frac{1}{\sqrt{6}}
\arctan \left( \frac{x}{\sqrt{6}} \right) + c \\
\amp = \frac{31}{2} \ln |x^2+6| + \frac{98}{\sqrt{6}}
\arctan \left( \frac{x}{\sqrt{6}} \right) + c
\end{align*}
Here is the complete integral.
\begin{align*}
\int \frac{2x^2+4x-1}{(x^2+3x+4)(x^2+6)} \amp = \frac{1}{58}
\left[ \int \frac{-31x - 75}{x^2+3x+4} dx + \int
\frac{31x+98}{x^2+6} dx \right] \\
\amp = \frac{1}{58} \left[ \frac{-31}{2} \ln |x^2+3x+4| -
\frac{57}{\sqrt{7}} \arctan \left( \frac{2x + 3}{\sqrt{7}}
\right) \right.\\
\amp \left. + \frac{31}{2} \ln |x^2+6| + \frac{98}{\sqrt{6}}
\arctan \left( \frac{x}{\sqrt{6}} \right) \right] + c \\
\amp = \frac{-31}{116} \ln |x^2+3x+4| -
\frac{57}{58\sqrt{7}} \arctan \left( \frac{2x + 3}{\sqrt{7}}
\right) \\
\amp + \frac{31}{116} \ln |x^2+6| + \frac{49}{29\sqrt{6}}
\arctan \left( \frac{x}{\sqrt{6}} \right) + c
\end{align*}
