find the integral of
f (x) = (1+x^2)ArcTan[ x]
My, where you do get these complicated integration from?
Here is what you do,
INTEGRAL (1+x^2)^2*atan(x)*(1+x^2)^{-1} dx
Let, u=atan(x) then, u'=(1+x^2)^{-1}
Then,
sec^2 u=1-x^2
cos^2 u=x^2
Then,
sec^2 u +2cos^2 u=1+x^2
Thus,
INTEGRAL (sec^2 u +2cos^2 u)^2 u*u' dx
Substitution rule it is composition of outer function,
INTEGRAL (sec^2 u+2cos^2 u)^2 du
Open parantheses, (secant*cosine=1)
INTEGRAL (sec^4 u+4cos^4 u+2) du
And this is definitely solvable
Hello, fair_lady0072002!
Let: u= arctan(x) . → . x = tan(u) . → . dx = sec²(u) duFind the integral of: .(1 + x²) arctan(x) dx
. . Then: .1 + x² .= .1 + tan²(x) .= .sec²(x)
Substitute: . ∫ sec²(u) · u · sec²(u) du . = . ∫ u·sec^4(u) du
. . . Good luck!