Question

Sol. [Key Concept: Split the integrand into odd and even parts: the odd part integrates to $0$ over $\lbrack - 1,1\rbrack$, leaving $2\int_{0}^{1}\frac{1}{x + 1}\ dx = 2ln2$.]

$${I = \int_{- 1}^{1}\mspace{2mu}\frac{x^{3} + |x| + 1}{x^{2} + 2|x| + 1}.dx }{I = \int_{- 1}^{1}\mspace{2mu}\frac{x^{3}}{x^{2} + 2|x| + 1} \cdot dx + \int_{- 1}^{1}\mspace{2mu}\frac{|x| + 1}{x^{2} + 2|x| + 1} \cdot dx }{I = 0 + \int_{- 1}^{1}\mspace{2mu}\frac{|x| + 1}{|x|^{2} + 2|x| + 1}.dx }{2\int_{0}^{1}\mspace{2mu}\frac{1}{x + 1} \cdot dx }{I = 2\lbrack ln|x + 1|\rbrack_{0}^{1} }{I = 2\mathcal{l}n(2)}$$