Problems on Techniques of Integration

The answer is given in any Table of integrals. Here we try to give the details for getting the answer. Since the degree of the numerator is bigger than the degree of the denominator, we will perform the long division to get

\begin{displaymath}\frac{x^n}{ax^2 + b x + c}= \frac{1}{a} x^{n-2} - \frac{1}{a}\;\;\frac{bx^{n-1} + cx^{n-2}}{ax^2 + b x + c}\end{displaymath}


\begin{displaymath}\int \frac{x^n}{ax^2 + b x + c}dx = \frac{1}{a(n-1)}x^{n-1} - \frac{1}{a}\; \int \frac{bx^{n-1} + cx^{n-2}}{ax^2 + b x + c}dx\end{displaymath}


\begin{displaymath}\int \frac{x^n}{ax^2 + b x + c}dx = \frac{1}{a(n-1)}x^{n-1} -...
...b x + c}dx - \frac{c}{a}\;\int \frac{x^{n-2}}{ax^2 + b x + c}dx\end{displaymath}

In other words, we will obtain the given integral by the above recurrent formula. Indeed, set

\begin{displaymath}F_n(x) = \int \frac{x^n}{ax^2 + b x + c}dx\end{displaymath}

then we have

\begin{displaymath}F_n(x) = \frac{1}{a(n-1)}x^{n-1} - \frac{b}{a} F_{n-1}(x) - \frac{c}{a} F_{n-2}(x)\cdot\end{displaymath}

It is then enough to know $F_0(x)$ and $F_1(x)$ to generate all of the functions $F_n(x)$.

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