Here,
I=inttan^-1x^2dx
=int(1)tan^-1x^2dx
"Using "color(blue)"Integration by parts":
color(blue)(int(uv)dx=uintvdx-int(u'intvdx)dx
Let, u=tan^-1x^2 and v=1
u'=1/(1+x^4)xx2x=(2x)/(1+x^4)andintvdx=x
I=tan^-1x^2(x)-int(2x)/(1+x^4)xxxdx
=xtan^-1x^2-int(2x^2)/(1+x^4)dx
I=xtan^-1x^2-I_A,where,...tototo(psi)
I_A=int(2x^2)/(1+x^4)dx
=int((x^2+1)+(x^2-1))/(1+x^4)dx
=int(x^2+1)/(x^4+1)dx+int(x^2-1)/(x^4+1)dx
=int(1+1/x^2)/(x^2+1/x^2)dx+int(1-1/x^2)/(x^2+1/x^2)dx
I_A=int(1+1/x^2)/((x-1/x)^2+2)dx+int(1-1/x^2)/((x+1/x)^2-2)dx
Take ,x-1/x=u,=>(1+1/x^2)dx=du, in first integral
and v=x+1/x,=>(1-1/x^2)dx=du ,in second integral
So,
I_A=int1/(u^2+(sqrt2)^2)du+int1/(v^2-(sqrt2)^2)dv
=1/sqrt2tan^-1(u/sqrt2)+1/(2sqrt2)ln|(v-sqrt2)/(v+sqrt2)|+c
=1/sqrt2tan^-1((x-1/x)/sqrt2)+1/(2sqrt2)ln|(x+1/x-
sqrt2)/(x+1/x+sqrt2)|+c
I_A=1/sqrt2tan^-1((x^2-1)/(sqrt2x))+1/(2sqrt2)ln|(x^2-
sqrt2x+1)/(x^2+sqrt2x+1)|+c
Hence, from (psi)
I=xtan^-1x^2-1/sqrt2tan^-1((x^2-1)/(sqrt2x))-1/(2sqrt2)ln|(x^2-
sqrt2x+1)/(x^2+sqrt2x+1)|+c
