Problems on Techniques of Integration

Since the derivative of $x^2 + x + 1$ is $2x+1$, and

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

we get

\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}

This implies

\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}

Also

\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}

Detailed Answer.


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