Answer
$\frac{1}{2} [(x^2+1)\operatorname{arccsc}(x^2+1) + \ln|x^2 +1 + \sqrt {x^4+2x^2}| ] +C$
Work Step by Step
Integrate by tables
First, fit the integrand to one of the integration tables
$\int x arccsc(x^2+1)dx$
Let $u=x^2+1$, and $du=2xdx$
$\frac{1}{2} \int arccsc(u)du$
Use formula #80 from the integration table and substitute u
$\frac{1}{2}[ (x^2+1) arccsc(x^2+1) + \ln |{(x^2+1) + \sqrt {(x^2+1)^2-1}}|] +C$
$\frac{1}{2} [(x^2+1)arccsc(x^2+1) + \ln(x^2 +1 + \sqrt {x^4+2x^2} ] +C$
