_-1^1((x^3+|x|+1)/(x^2+2|x|+1))dx is equal to:

Question

$\displaystyle\int_{-1}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)dx$ is equal to:

Accepted Answer

Split at $x=0$. For $-1\le x<0$: $|x|=-x$, integrand $=\dfrac{x^3-x+1}{(x-1)^2}$. Polynomial division gives $x+2+\dfrac{2}{x-1}+\dfrac{1}{(x-1)^2}$.

$$I_1 = \int_{-1}^0\left(x+2+\frac{2}{x-1}+\frac{1}{(x-1)^2}\right)dx = \left[\frac{x^2}2+2x+2\ln|x-1|-\frac1{x-1}\right]_{-1}^0 = 2-2\ln2$$

For $0\le x\le 1$: $|x|=x$, integrand $=\dfrac{x^3+x+1}{(x+1)^2}$, giving $x-2+\dfrac{4}{x+1}-\dfrac{1}{(x+1)^2}$.

$$I_2 = \int_0^1\left(x-2+\frac{4}{x+1}-\frac{1}{(x+1)^2}\right)dx = \left[\frac{x^2}2-2x+4\ln|x+1|+\frac1{x+1}\right]_0^1 = -2+4\ln2$$

$$I = I_1+I_2 = (2-2\ln2)+(-2+4\ln2) = 2\ln 2$$

Solutions · 1