The value of the integral is equal to: [2026]

Question

The value of the integral $\int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$ is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

$f (x) = \int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$

$f (x) = \int_{- 1}^{1} \frac{x^{3}}{x^{2} + 2 | x | + 1} d x + \int_{}^{} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$

Since, $\frac{x^{3}}{x^{2} + 2 | x | + 1}$ is an odd function and $\frac{| x | + 1}{x^{2} + 2 | x | + 1}$ is an even function

$∴ f (x) = 2 \int_{0}^{1} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$

$= 2 \int_{0}^{1} \frac{x + 1}{(x + 1)} d x$

$= 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 (\log_{e} {(x + 1))}_{0}^{1} = 2 (\log_{e} 2)$

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