How to integrate $ \int\frac{x^3+|x|+1}{x^2+2|x|+1} $?

Question

$$ \int\frac{x^3+|x|+1}{x^2+2|x|+1} $$

I tried redefining the function , but that could not help . I even tried making factors out of denominator which again proved futile .

Any help to solve this integral would be admired !

Answer 1

Hint:

For real $x$

If $x\ge0,|x|=+x, x^2+2|x|+1=(x+1)^2,$ choose $x+1=y$

If $x<0,|x|=-x,$ choose $x-1=y$

Answer 2

Write down the integrand separately for the regions of positive and negative $x$

$$f(x\lt0) = f_{m}(x) = \frac{x^3-x+1}{x^2-2 x+1}$$

$$f(x\gt0) = f_{p}(x) = \frac{x^3+x+1}{x^2+2 x+1}$$

Now integrate $f(x)$ over each region, taking care of continuity at $x=0$ which, up to an added constant, gives the function $i(x)$ asked for in the OP:

$$i(x\lt0)= i_{m}(x) = \int_{-1}^{x}f_{m}(t)\,dt \\= \frac{1}{2(x-1)}\left(x^3+3 x^2-x (2+\log (16))\\+4 (x-1)\log (1-x)-4+\log(16)\right)$$

$$i(x\gt0) = i_{p}(x) = \int_{0}^{x}f_{p}(t)\,dt + i_{m}(0) \\= \frac{x ((x-3) x-6)}{2 (x+1)}+4 \log (x+1)+\frac{1}{2} (4-\log (16))$$