Question

Evaluate the following integral:

\[\int_{- 2}^2 \frac{3 x^3 + 2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]

Answer 1

\[\text{Let I} = \int_{- 2}^2 \frac{3 x^3 + 2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]

\[= \int_{- 2}^2 \frac{3 x^3}{x^2 + \left| x \right| + 1}dx + \int_{- 2}^2 \frac{2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]
\[ = I_1 + I_2\]

Consider

\[f\left( x \right) = \int_{- 2}^2 \frac{3 x^3}{x^2 + \left| x \right| + 1}dx\]

\[f\left( - x \right) = \int_{- 2}^2 \frac{3 \left( - x \right)^3}{\left( - x \right)^2 + \left| - x \right| + 1}dx = \int_{- 2}^2 \frac{- 3 x^3}{x^2 + \left| x \right| + 1}dx = - \int_{- 2}^2 \frac{3 x^3}{x^2 + \left| x \right| + 1}dx = - f\left( x \right)\]

\[\therefore I_1 = \int_{- 2}^2 \frac{3 x^3}{x^2 + \left| x \right| + 1}dx = 0 ......................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

Now, consider

\[g\left( x \right) = \int_{- 2}^2 \frac{2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]

\[g\left( - x \right) = \int_{- 2}^2 \frac{2\left| - x \right| + 1}{\left( - x \right)^2 + \left| - x \right| + 1}dx = \int_{- 2}^2 \frac{2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx = g\left( x \right)\]

\[\therefore I_2 = \int_{- 2}^2 \frac{2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx \]
\[ = 2 \int_0^2 \frac{2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx ..................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]
\[ = 2 \int_0^2 \frac{2x + 1}{x^2 + x + 1}dx .................\left[ \left| x \right| = \begin{cases}x, & x \geq 0 \\ - x, & x < 0\end{cases} \right]\]
\[ = \left.2 \times \log\left( x^2 + x + 1 \right)\right|_0^2 ....................\left[ \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log f\left( x \right) + C \right]\]
\[ = 2 \times \left( \log7 - \log1 \right)\]
\[ = 2 \times \left( \log7 - 0 \right)\]
\[ = 2\log7\]

\[\therefore I = I_1 + I_2 = 0 + 2\log7 = 2\log7\]