$\int\limits_{-1}^1\frac{x^3 + |x| + 1}{x^2+2|x|+1} dx$ is equal to

$2\log_e2$

The correct answer is Option (2) → $2\log_e2$

Evaluate

$\displaystyle I=\int_{-1}^{1}\frac{x^{3}+|x|+1}{x^{2}+2|x|+1}\,dx$

Split at $0$ and use $|x|=x$ for $x\ge0$, $|x|=-x$ for $x<0$:

$\displaystyle I=\int_{-1}^{0}\frac{x^{3}-x+1}{(1-x)^{2}}\,dx+\int_{0}^{1}\frac{x^{3}+x+1}{(1+x)^{2}}\,dx$

In the first integral substitute $x=-t$ (so $dx=-dt$):

$\displaystyle \int_{-1}^{0}\frac{x^{3}-x+1}{(1-x)^{2}}\,dx =\int_{1}^{0}\frac{-t^{3}+t+1}{(1+t)^{2}}(-dt) =\int_{0}^{1}\frac{-t^{3}+t+1}{(1+t)^{2}}\,dt$

Therefore

$\displaystyle I=\int_{0}^{1}\frac{-t^{3}+t+1}{(1+t)^{2}}\,dt+\int_{0}^{1}\frac{t^{3}+t+1}{(1+t)^{2}}\,dt =\int_{0}^{1}\frac{2t+2}{(1+t)^{2}}\,dt$

Simplify integrand:

$\displaystyle \frac{2t+2}{(1+t)^{2}}=\frac{2(1+t)}{(1+t)^{2}}=\frac{2}{1+t}$

Integrate:

$\displaystyle I=2\int_{0}^{1}\frac{1}{1+t}\,dt=2\big[\ln(1+t)\big]_{0}^{1}=2\ln 2$

Final answer: $I=2\ln 2$