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 denominator is 2 with an irreducible polynomial function. So no need to perform the long-division. One may also use the table for this integration.

First we will get ridd of the term $x$ in the numerator by using the derivative of the denominator. Indeed, we have

\begin{displaymath}x-1 = \frac{1}{2} \Big(2x+1\Big) - \frac{3}{2}\end{displaymath}

which implies

\begin{displaymath}\frac{x-1}{x^2 + x + 1} = \frac{1}{2} \frac{2x+1}{(x^2 + x + 1)} - \frac{3}{2}\frac{1}{(x^2 + x + 1)}\cdot\end{displaymath}

Hence

\begin{displaymath}\int \frac{x-1}{x^2 + x + 1}dx = \frac{1}{2} \int \frac{2x+1}{x^2 + x + 1}dx - \frac{3}{2} \int \frac{1}{x^2 + x + 1}dx\cdot\end{displaymath}

On the other hand, we have

\begin{displaymath}\int \frac{2x+1}{x^2 + x + 1}dx = \ln(x^2 + x + 1).\end{displaymath}

In order to integrate $\displaystyle \frac{1}{x^2 + x + 1}$, we will complete the square in the denominator and then use the $\arctan$ formula to get

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

We get

\begin{displaymath}\int \frac{x-1}{x^2 + x + 1}dx = \frac{1}{2} \ln(x^2 + x + 1) - \sqrt{3} \arctan\left(\frac{2x+1}{\sqrt{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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