I love Integration. (1 Viewer)

[Gussy Booo]

∫x.arctan²x dx.

Someone do it ^.^. Thanks in advanced !

The answers here --> http://integrals.wolfram.com/index.jsp?expr=x+%28arctan%28x%29%29^2&random=false

 

[tommykins]

try integration by parts u = (atanx)^2 and v' = x

do it twice and it should come out somewhere

 

[Aquawhite]

I get really close but couldn't quite get it. I will give it another go later. This is an insanely difficult question.

 

[Aquawhite]

Here: Solution (working backwards)

If you work backwards and try to understand all these steps, it should help. It has a lot more steps than I was willing to go to! I went so far and got stuck.

Just click show steps.

 

[Gussy Booo]

Lmfaooooooooo.
This won't help >.<

I found the solution. Here x)

∫ x arctan²(x) dx

you need to solve this one by parts

u = arctan(x)
arctan(x) = u
x = tan(u)
dx = sec²(u) du
dx = (1 + tan²(u)) du
dx = (1 + x²) du
du = dx/(1 + x²)

-----------------
dv = x arctan(x) dx
now this also needs to be solved by parts

u = arctan(x)
du = dx/(1 + x²)
dv = xdx
v = ½ x²

∫ x arctan(x) dx
= ½ x² arctan(x) - ½ ∫x² dx / (1 + x²)

for the second bit use the trig sub x = tan(t)
dx = sec²(t) dt

= ½ x² arctan(x) - ½ ∫tan²(t) sec²(t) dt / (1 + tan²(t))
= ½ x² arctan(x) - ½ ∫tan²(t) sec²(t) dt / sec²(t)
= ½ x² arctan(x) - ½ ∫tan²(t) dt
= ½ x² arctan(x) - ½ ∫sec²(t) - 1 dt
= ½ x² arctan(x) - ½ (tan(t) - t)
= ½ [x² arctan(x) - x + arctan(x)]
= ½ [(x² + 1)arctan(x) - x]
-----------------

so in summary
u = arctan(x)
du = dx/(1 + x²)
dv = x arctan(x) dx
v = ½ [(x² + 1)arctan(x) - x]

so by parts:

∫ x arctan²(x) dx
= ½ arctan(x)[(x² + 1)arctan(x) - x] - ∫ ½ [(x² + 1)arctan(x) - x] dx/(1 + x²)
= ½ [(x² + 1)arctan²(x) - x arctan(x) - ∫ (x² + 1)arctan(x)dx/(1 + x²) + ∫ x dx/(1 + x²)]
= ½ [(x² + 1)arctan²(x) - x arctan(x) - ∫arctan(x) dx + ∫ x dx/(1 + x²)]

-----------------
lets look at
∫arctan(x) dx

do it by parts
u = arctan(x)
du = dx/(1 + x²)
dv = dx
v = x

∫arctan(x) dx = x arctan(x) - ∫x dx / (1 + x²)

------------------
= ½ [(x² + 1)arctan²(x) - x arctan(x) - x arctan(x) + ∫x dx / (1 + x²) + ∫ x dx/(1 + x²)]
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + 2∫x dx / (1 + x²)]

.........
let x = tan(t)
dx = sec²(t) dt
.........
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + 2∫ tan(t) sec²(t) dt / (1 + tan²(t))]
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + 2∫ tan(t) sec²(t) dt / sec²(t)]
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + 2∫ tan(t) dt]
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + 2∫sin(t) dt/cos(t)]
....
let u = cos(t)
du = -sin(t)dt
....

= ½ [(x² + 1)arctan²(x) - 2x arctan(x) - 2∫ du/u]
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) - 2ln|u|] + C
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) - 2ln|cos(t)|] + C
..........
x = tan(t)
x² = tan²(t)
x² = sec²(t) - 1
x² + 1 = sec²(t)
sec(t) = √(x² + 1)
cos(t) = 1/√(x² + 1)
...............

= ½ [(x² + 1)arctan²(x) - 2x arctan(x) - 2ln|cos(t)|] + C
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) - 2ln(1/√(x² + 1))] + C
= ½ [(x² + 1)arctan²(x) - 2x arctan(x) + ln(x² + 1)] + C
= ½ (x² + 1)arctan²(x) - x arctan(x) + ½ ln(x² + 1) + C

 

[Aquawhite]

If only you could have done it all on your own

 

[jet]

No need for trig substitutions at all

 

[Aquawhite]

I love you jet!