Question

How do you evaluate `int (x*arctanx) dx / (1 + x^2)^2` from 0 to infinity?

Answer 1

`int_0^{oo} (x*arctanx) dx / (1 + x^2)^2 = pi/8`

Explanation:

`d/(dx)(arctan(x)/(1+x^2)) =1/(1 + x^2)^2 - (2 x arcTan(x))/(1 + x^2)^2`

so

`int (x*arctanx) dx / (1 + x^2)^2 = 1/2int dx/(1 + x^2)^2- 1/2arctan(x)/(1+x^2)`

but

`1/(1 + x^2)^2 = 1/2 d/(dx)(x/(1 + x^2))+1/2 d/(dx) (arctan(x))`

concluding

`int (x*arctanx) dx / (1 + x^2)^2=(x + (x^2-1) arcTan(x))/(4 (1 + x^2))`

but

`lim_{x->0}(x + (x^2-1) arcTan(x))/(4 (1 + x^2)) = 0`

and

`lim_{x->oo}(x + (x^2-1) arcTan(x))/(4 (1 + x^2)) = (pi/2)/4 = pi/8`

finally

`int_0^{oo} (x*arctanx) dx / (1 + x^2)^2 = pi/8`

Answer 2

Let `I=int_0^ oo(x*arctanx)/(1+x^2)^2dx.`

We take sbstn., `arctanx=t`, so that, `x=tant`, and, `dx=sec^2t.`

Also, `x=0 rArr t=0,` and `x rarr oo, trarr pi/2.`

Because of all these changes, now, `I` becomes,

`I=int_0^(pi/2)(t*tant*sec^2t)/sec^4tdt=int_0^(pi/2)(t*tant)/sec^2tdt`

`=int_0^(pi/2)(t)(sint/cost)cos^2tdt=int_0^(pi/2)tsintcostdt`

`=(1/2)int_0^(pi/2)tsin2tdt.................(i)`

`=1/2[t*{-cos(2t)/2}]_0^(pi/2)-(1/2)int_0^(pi/2)[1*{-cos(2t)/2}]dt`
`=-1/4[{pi/2*cospi}-0]+1/4[sin(2t)/2]_0^(pi/2)`
`=pi/8+1/8[sinpi-sin0]`
`:. I=pi/8`.

To evaluate `I` further from `(i)`, we have used the Rule of Integration by Parts for Definite Integral :

`int_a^bu*vdx=[u*intvdx]_a^b-int_a^b{(du)/dx*intvdx}dx,` with `u=t, &, v=sin2t.`

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Answer 3

`pi/8`

Explanation:

Use the substitution `x = tan u and 1+tan^2u=sec^2 u`..

The limits become `pi/2 and oo, and dx = sec^2 u` `du`.

The given integral reduces to the form

`int u sin u cos u du=1/2intu sin 2u du`

`=(-1/4)intud(cos 2u)`

`=-1/4[u cos 2u-int cos 2u du]`, between the limits `0 and pi/2`

=. `(-1/4)(pi/2)(-1)+1/4[1/2sin 2u]`, between the limits

`=pi/8+0=pi/8`

-

Answer 4

`= pi/8`

Explanation:

`int_0^oo dx qquad (x*arctanx) / (1 + x^2)^2`

using IBP

`u = arctan x, u' = 1/(1+ x^2)`

`v' = x / (1 + x^2)^2, v = -1 /(2 (1 + x^2))`

`arctan x (-1 /(2 (1 + x^2)) ) - int_0^oo dx qquad 1/(1+ x^2)* (-1 /(2 (1 + x^2)))`

`= -arctan x /(2 (1 + x^2)) + 1/2 color{red}{int_0^oo dx qquad 1/( (1 + x^2)^2)} qquad star`

for the red bit we use `x = tan psi, dx = sec^2 psi d psi`

`implies int dpsi qquad sec^2 psi / sec^4 psi`

`= int dpsi qquad cos^2 psi`

`= 1/2 int dpsi qquad cos2 psi +1`

`= 1/2 ( 1/2 sin 2 psi + psi)`

`= 1/2 ( sin psi cos psi + psi)`

`= 1/2 ( x/sqrt{1 + x^2} * 1/sqrt{1 + x^2} + arctan x)`

`= 1/2 ( x/(1 + x^2) + arctan x)`

subbing back into `star`

`= [ -arctan x /(2 (1 + x^2)) + 1/2 (1/2 ( x/(1 + x^2) + arctan x)) ]_0^oo`

`= [ -arctan x /(2 (1 + x^2)) + 1/4 * x/(1 + x^2) + 1/4 arctan x ]_0^oo`

`= pi/8`