I have to calculate this integralsI1=∫02arctan⁡(x)x2+2x+2dxI2=limn→∞∫0narctan⁡xx2+x+1dxI have a hint on the first one, the substitution x=2−t1+2t, but I can't understand how it was conceived.

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I have to calculate this integralsI1=∫02arctan⁡(x)x2+2x+2dxI2=limn→∞∫0narctan⁡xx2+x+1dxI have a hint on the first one, the substitution x=2−t1+2t, but I can&#039;t understand how it was conceived.

Bukvald5z · Accepted Answer

I2=∫0+∞arctan⁡xx2+x+1dx=∫01arctan⁡xx2+x+1dx+∫1+∞arctan⁡xx2+x+1dx  In the second integral perform the change of variable y=1x  I2=∫01arctan⁡xx2+x+1dx+∫01arctan⁡(1x)x2+x+1dx  =∫01(arctan⁡x+arctan⁡(1x))x2+x+1dx  =∫01π2x2+x+1dx  =π2∫011(x+12)2+34dx  =π2×23[arctan⁡(2x+13)]01  =π3(arctan⁡(3)−arctan⁡(13))  =π3(π3−π6))  =π263  Other important relation for arctan function:  If x&amp;gt;0 then,  arctan⁡x+arctan⁡(1x)=π2  If x&amp;lt;0 replace π2 by −π2  And remember too that, for all x real,

Esther Phillips · Accepted Answer

In I1 apply the change of variable x=2−t1+2t  Observe that t=2−x1+2x  I1=∫02arctan⁡(2−t1+2t)t2+2t+2dt  If t∈[0;2] then 0&amp;lt;2t&amp;lt;1 therefore,  I1=∫02arctan⁡(2)t2+2t+2dt−I1  Therefore,  I1=12∫02arctan⁡(2)t2+2t+2dt  =arctan⁡(2)2[arctan⁡(x+1)]02  =arctan⁡(2)arctan⁡(3)2−arctan⁡(2)8π

karton · Accepted Answer

∫02arctan⁡(x)x2+2x+2dx=I∫02ln⁡(1+xi)x2+2x+2dx(set t=1+ix⇒x=[1−t]i)R=∫i1+2iln⁡(t)(t+i)(t−2+i)dt=12R∫11+2iln⁡(t)−i−tdt−12R∫i1+2iln⁡(t)2−i−tdtHowever∫ln⁡(t)z−tdt=∫ln⁡(zr)1−rdr=−ln⁡(1−r)ln⁡(zr)+∫ln⁡(1−r)rdr=−ln⁡(1−r)ln⁡(zr)−Li2(r)=−ln⁡(1−tz)ln⁡(t)−Li2(tz)With equation becomes:∫02arctan⁡(x)x2+2x+2dx=12R[−ln⁡(1−1+2i−i)ln⁡(1+2i)−Li2(1+2i−i)+Li2(1−i)+ln⁡(1−1+2i2−iln⁡(1+2i)+Li2(1+2i2−i)−Li2(12−i)]=18[π−4arccot(3)]arctan⁡(2)−18ln2⁡(5)−12RLi2(−2+i)+12RLi2(i)+12RLi2(i)−12RLi2(2+i5)18[π−4arccot(3)]arctan⁡(2)−18ln2⁡(5)−π248−12RLi2(−2+i)−12RLi2(2+i5) ≈0.2567

I have to calculate this integrals I_1=\int_0^2\frac{\arctan(x)}{x^2+2x+2}dx I_2=\lim_{n\to\infty}\int_0^n\frac{\arctan x}{x^2+x+1}dx I have a hint

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New Questions in Calculus 1