How do I evaluate the definite integral of [x arctan(x^2)] over the interval (1,0)?

Question

I know that the answer is pi/8 -1/4ln(2) and that this answer is obtained by means of integration by parts(or maybe not). However, I can't figure out how to arrive at this answer. Can someone please give me some direction? Thanks.

Answer 1

trying...
IBP yes.
Let u = arctan(x²) ; du = 2x/(1+x^4)dx
dv = x dx ; v = ½x²

∫ x arctan(x²) dx = ½x²arctan(x²) - ∫x³/(1+x^4)dx
= ½x²arctan(x²) - ¼ln (1+x^4)

at x = 0, everything is zero
at x = 1, it is ½arctan(1) - ¼ln2 = π/8 - ¼ln2.
That is also the answer when the definite integral is evaluated from 0 to 1.

Answer 2

Let u = x², du = 2x dx.
When x = 0, u = 0 and when x = 1, u = 1
So we have to compute
1/2 *∫(0..1) arctan u du.
We do ∫(0..1) arctan u du
by parts:
Let U = arctan u dV = du
dU = du/(1+u²) V = u
So ∫ (0..1)arctan u du = u arctan u(0..1) - ∫(0..1) u du/(1+u²)
= π/4 - ln(2)/2.
Since there was a 1/2 in front of the substituted integral,
the final answer is π/8 - ln(2)/4.

Answer 3

let us put x^2 = t

the xdx = dt/2

at x = 1 t = 1 and x= 0 t = 0

so we need integral (arc tan t)/2 from 1 to 0

you should be able to proceed