Answer
$$\int\frac{x^3+x+1}{(x^2+1)^2}dx=\frac{1}{4}\ln(x^4+2x^2+1)+\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x+C$$
Or:
$$=\frac{1}{2}\ln(x^2+1)+\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x+C$$
Work Step by Step
$$A=\int\frac{x^3+x+1}{(x^2+1)^2}dx$$ $$A=\int\frac{x^3+x}{(x^2+1)^2}dx+\int\frac{1}{(x^2+1)^2}dx$$ $$A=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{1}{(x^2+1)^2}dx$$
For the first integral, take $u=x^4+2x^2+1$, which makes
$$du=(4x^3+4x)dx$$ $$(x^3+x)dx=\frac{1}{4}du$$
For the second integral, use Formula 33, which states that
$$\int\frac{dx}{(a^2+x^2)^2}=\frac{x}{2a^2(a^2+x^2)}+\frac{1}{2a^3}\tan^{-1}\frac{x}{a}+C$$
for $a=1$ here
$$A=\frac{1}{4}\int\frac{1}{u}du+\Big[\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x\Big]+C$$ $$A=\frac{1}{4}\ln|u|+\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x+C$$ $$A=\frac{1}{4}\ln|x^4+2x^2+1|+\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x+C$$
As $x^4+2x^2+1\gt0$ for all $x$:
$$A=\frac{1}{4}\ln(x^4+2x^2+1)+\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}x+C$$
