How do you evaluate the integral #int x^-2arcsinx#?

Use integration by parts. Let:

#{(u=arcsinx,=>,du=1/sqrt(1-x^2)dx),(dv=x^-2dx,=>,v=-x^-1):}#

Then:

#intx^-2arcsinxdx=-arcsinx/x+int1/(xsqrt(1-x^2))dx#

Just working with the remaining integral, let #x=sintheta#. This implies that #sqrt(1-x^2)=costheta# and #dx=costhetad theta#. Then:

#int1/(xsqrt(1-x^2))dx=int1/(sinthetacostheta)costhetad theta=intcscthetad theta#

Which is a commonly known integral. Also note that if #sintheta=x#, this is represented in a right triangle where the side opposite #theta# is #x# and the hypotenuse is #1#. The leg adjacent to #theta#, then, is #sqrt(1-x^2)#. From this right triangle we can say that #csctheta=1/x# and #cottheta=sqrt(1-x^2)/x#, which will be relevant:

#int1/(xsqrt(1-x^2))dx=-lnabs(csctheta+cottheta)=-lnabs(1/x+sqrt(1-x^2)/x#

Combining the denominators and inverting the fraction by bringing the #-1# outside the natural log inside as a #-1# power, and putting this result into the original expression we found from integration by parts, we find a final answer of:

#intx^-2arcsinxdx=lnabs(x/sqrt(1-x^2))-arcsinx/x+C#

Let #x=sintheta#. This implies that #dx=costhetad theta# and that #theta=arcsinx#. Then:

#I=intx^-2arcsinxdx=intcsc^2theta(theta)(costhetad theta)=intthetacotthetacscthetad theta#

Then perform integration by parts. Let:

#{(u=theta,=>,du=d theta),(dv=cotthetacscthetad theta,=>,v=-csctheta):}#

So:

#I=uv-intvdu=-thetacsctheta+intcscthetad theta#

Which is a common integral:

#I=-thetacsctheta-lnabs(csctheta+cottheta)=(-theta)/sintheta-lnabs((1+costheta)/sintheta)#

Rewriting the natural logarithm by bringing the #-1# into the integral as a #-1# power:

#I=(-theta)/sintheta+lnabs(sintheta/(1+sqrt(1-sin^2theta))#

And since #sintheta=x#:

#I=lnabs(x/(1+sqrt(1-x^2)))-arcsinx/x+C#