What is the integration of int((x(tanx)^-1)dx)/(1+x^2)^(3/2) ?

int((x(tanx)^-1)dx)/(1+x^2)^(3/2)

1 Answer
Jul 30, 2018

int((x(tanx)^-1)dx)/(1+x^2)^(3/2)=-(tan^-1x)(1/sqrt(1+x^2))+x/sqrt(1+x^2)+C

Explanation:

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