Problems on Techniques of Integration

We have

$$\frac{x^4 + 2x^3 + 3x^2 -3 x -8}{x^2 + 2 x +5} = x^2 - 2 + \frac{x+2}{x^2 + 2 x +5}\cdot$$

So

$$\int \frac{x^4 + 2x^3 + 3x^2 -3 x -8}{x^2 + 2 x +5}dx = \frac{x^3}{3} -2 x + \int \frac{x+2}{x^2 + 2 x +5} dx \cdot$$

In order to integrate $$\frac{x+2}{x^2 + 2 x +5}$$, we use the identity

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

remembering the derivative of $x^2 + 2x + 5$ being $2x+2$. So

$$\frac{x+2}{x^2 + 2x + 5} = \frac{1}{2} \frac{2x+2}{(x^2 + 2x + 5)} + \frac{1}{(x^2 + 2x + 5)}\cdot$$

This implies

$$\int \frac{x+2}{x^2 + 2x + 5}dx = \frac{1}{2} \int \frac{2x+2}{x^2 + 2x + 5}dx + \int \frac{1}{x^2 + 2x + 5}dx\cdot$$

On one hand, we have

$$\int \frac{2x+1}{x^2 + 2x + 5}dx = \ln(x^2 + 2x + 5),$$

and also

$$\int \frac{1}{x^2 + 2x + 5}dx = \int \frac{1}{(x + 1)^2 + 4}dx = \frac{1}{2}\arctan\left(\frac{x+1}{2}\right).$$

We get

$$\int \frac{x+2}{x^2 + 2x + 5}dx = \frac{1}{2} \ln(x^2 + 2x + 5) +\frac{1}{2}\arctan\left(\frac{x+1}{2}\right) + C.$$

Detailed Answer.

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