Problems on Techniques of Integration

We recognize here a rational function. The technique of integrationg such functions is the partial decomposition technique. In this case, the degree of the numerator is bigger than the degree of the denominator. So we need to perform the long-division. We get

\begin{displaymath}\frac{3 x^6 -2 x^5 + 32 x^4 -16 x^3 + 42x^2 + 18 x -27}{x^2 + 9} = 3x^4-2x^3+5x^2+2x-3 \;+\; \frac{x+1}{x^2 + 9}\cdot\end{displaymath}


\begin{displaymath}\int \frac{3 x^6 -2 x^5 + 32 x^4 -16 x^3 + 42x^2 + 18 x -27}{...
... + \frac{5x^3}{3}+ x^2 - 3x + \int \frac{x+1}{x^2 + 9} dx \cdot\end{displaymath}

In order to integrate $\displaystyle \frac{x+1}{x^2 + 9}$, we will first get ridd of the term $x$ in the numerator by using the derivative of the denominator. Indeed, we have

\begin{displaymath}\frac{x+1}{x^2 + 9} = \frac{1}{2} \frac{2x}{(x^2 + 9)} + \frac{1}{x^2 + 9}\cdot\end{displaymath}


\begin{displaymath}\int \frac{x+1}{x^2 + 9}dx = \frac{1}{2} \int \frac{2x}{x^2 + 9}dx + \int \frac{1}{x^2 + 9}\cdot\end{displaymath}

On one hand, we have

\begin{displaymath}\int \frac{2x}{x^2 + 9}dx = \ln(x^2 + 9),\end{displaymath}

and also

\begin{displaymath}\int \frac{1}{x^2 + 9}dx = \frac{1}{3}\arctan\left(\frac{x}{3}\right).\end{displaymath}


\begin{displaymath}\int \frac{x+1}{x^2 + 9}dx = \frac{1}{2} \ln(x^2 + 9) +\frac{1}{3}\arctan\left(\frac{x}{3}\right) + C.\end{displaymath}

It is a common mistake to forget the constant $C$.

Remark. Recall the $\arctan$ formula

\begin{displaymath}\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left(\frac{x}{a}\right)\end{displaymath}

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