Problems on Techniques of Integration

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

$$x-1 = \frac{1}{2} \Big(2x+1\Big) - \frac{3}{2}$$

we get

$$\frac{x-1}{x^2 + x + 1} = \frac{1}{2} \frac{2x+1}{(x^2 + x + 1)} - \frac{3}{2}\frac{1}{(x^2 + x + 1)}\cdot$$

This implies

$$\int \frac{x-1}{x^2 + x + 1}dx = \frac{1}{2} \int \frac{2x+1}{x^2 + x + 1}dx - \frac{3}{2} \int \frac{1}{x^2 + x + 1}dx\cdot$$

On the other hand, we have

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

Also

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

We get

$$\int \frac{x-1}{x^2 + x + 1}dx = \frac{1}{2} \ln(x^2 + x + 1) - \sqrt{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right) + C.$$

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