int((x(tanx)^-1)dx)/(1+x^2)^(3/2)
Put x=tanz
dx=sec^2z
int (tanz.z.(sec^2z))/((sec^2z)^(3/2)).dz
int(tanz.z.dz)/secz.dz
int(sinz/cosz).cosz.z.dz
intsinz.z.dz
Now I'm gonna use integration by parts
intu.dv = u.v -intdu.v
u = z dz
du = 1
dv=sinz
v=-cosz
Now put values in formula
intz.sinz.dz=z(-cosz)-int(-cosz)
intz.sinz.dz=z(-cosz)+intcosz
intz.sinz.dz=z(-cosz)+sinz +C.......(1)
As we know cosz=1/secz
secz=sqrt(1+tan^2z)
And is equal to tanz =x above.
then, secz=sqrt(1+x^2)
cosz=1/sqrt(1+x^2)
sinz=sqrt(1-cos^2z)
sinz=sqrt(1-1/(1+x^2))
sinz=sqrt(x^2/(1+x^2))
sinz = x/sqrt(1+x^2)
Now put value of z, cosz, sinz in eq 1
Here is Final answerint((x(tanx)^-1)dx)/(1+x^2)^(3/2)=tan^-1x(-1/sqrt(1+x^2))+x/sqrt(1+x^2) +C