integral of xarctanxdx using integration by parts.

Answer 1

∫x arctan x dx

u = arctan x
du = 1/(1+x^2)

dv = x dx
v = 1/2 x^2

∫ x arctan x dx = 1/2 x^2 arctan x - ∫ 1/2 x^2 * 1/(1+x^2) dx
= 1/2 (x^2 arctan(x) - ∫x^2/(1+x^2) dx)

now, x^2/(1+x^2) = 1 - 1/(1+x^2)

∫x^2/(1+x^2) dx = ∫ 1 - 1/(1+x^2) dx
= x - arctan(x)

so we wind up with

1/2 (x^2 arctan(x) - (x - arctan(x)))
= 1/2 ((1+x^2)arctan(x) - x) + C

Answer 2

∫xarctan(x) dx

Integration by parts:

∫vdu = vu - ∫udv

Where:
v = arctan(x) du = xdx
dv = 1/(x^2 + 1)dx u = x^2/2

∫xarctan(x)dx = (x^2/2)arctan(x) - ∫(x^2)/2(x^2 + 1)dx

Factor out constants:

(1/2)x^2arctan(x) - (1/2)∫x^2/(x^2+1)dx

Long division in the integrand:

(1/2)x^2arctan(x) - (1/2)∫(1-(1/(x^2+1))dx

Separate integrand:

(1/2)x^2arctan(x) - (1/2)∫1dx -(1/2)∫1/(x^2+1)dx

Integrate:

(1/2)x^2arctan(x) - (x/2) - (1/2)arctan(x) + C

Factor:

(1/2)(x^2arctan(x) - x - arctan(x)) + C

Complete:

∫xarctan(x)dx = (1/2)(x^2arctan(x)-x-arctan(x)) + C

Answer 3

thank you so much! I was getting stuck at the long division part.