Answer
$$\int \frac{dx}{x(x^{4}+1)}=\frac{1}{4}ln(\frac{x^{4}}{x^{4}+1})+C$$
Work Step by Step
$$\int \frac{dx}{x(x^{4}+1)}=\int \frac{x}{x^{2}(x^{4}+1)}dx$$
$$let\ t=x^{2},\,dt=2x\,dx$$
$$\int \frac{x}{x^{2}(x^{4}+1)}dx=\frac{1}{2}\int \frac{1}{t(t^{2}+1)}dt$$
$$=\frac{1}{2}\int \frac{1+t^{2}-t^{2}}{t(t^{2}+1)}dt=\frac{1}{2}\int (\frac{1}{t}-\frac{t}{t^{2}+1})dt$$
$$=\frac{1}{2}ln\left | t \right |-\frac{1}{4}ln(t^{2}+1)+C$$
$$=\frac{1}{2}ln( x^{2})-\frac{1}{4}ln(x^{4}+1)+C$$
$$=\frac{1}{4}ln(\frac{x^{4}}{x^{4}+1})+C$$