What is #int arctan(x^2) dx#?

Question

Charles Arocha · Accepted Answer

#I=xtan^-1x^2-1/sqrt2tan^-1((x^2-1)/(sqrt2x))-1/(2sqrt2)ln|(x^2-sqrt2x+1)/(x^2+sqrt2x+1)|+c#

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#

Christopher Allers · Answer

undefined The integral of arctan(x^2) with respect to x is not expressible in terms of elementary functions. It cannot be represented by a combination of polynomials, exponential functions, logarithmic functions, trigonometric functions, or their inverses. Therefore, it is considered a non-elementary integral.

Aurora Alvarado · Answer

undefined The integral of $ \arctan(x^2) $ with respect to $ x $ is not expressible in terms of elementary functions. It cannot be represented by a combination of polynomials, exponentials, logarithms, trigonometric functions, or their inverses. Therefore, the integral $ \int \arctan(x^2) \, dx $ does not have a simple analytical solution.