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Anyone think they can tackle this question?

Having a little bit of trouble.

 May 2, 2017
edited by TakahiroMaeda  May 2, 2017
 #1
avatar+9488 
+4

          

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 May 2, 2017
 #3
avatar+118710 
+1

I love your illustrated thought process Hectictar :)

Melody  May 2, 2017
 #4
avatar+9488 
+1

Haha thank you Melody!

I obviously didn't make this though! Just thought I'd clarify that laughlaugh

hectictar  May 2, 2017
 #2
avatar+118710 
+2

I used this site:

https://goo.gl/Qem530

and got this answer:

I haven't tried to make sense of it yet though.

 

 May 2, 2017
 #5
avatar+26397 
+4

Something Mathematical to do.

x=01(x+1+x2)2 dxsubstitute: x=tan(z)dx=(1+tan2(z)) dz new limits: z=arctan(0)z=0z=arctan()z=π2 =π2z=01+tan2(z)(tan(z)+1+tan2(z))2 dz 1+tan2(z)=1cos2(z) =π2z=01cos2(z)(11cos2(z)+tan2(z)+2sin(z)cos2(z)) dz=π2z=011+2sin(z)+sin2(z) dz=π2z=01(1+sin(z))2 dzsubstitute: t=tan(z2)dt=12(1+tan2(z2)) dz=12(1+t2) dz new limits: t=tan(02)t=0t=tan(π22)t=1 =1t=02 dt1+t2(1(1+sin(z))2) sin(z)=2t1+t2 =1t=02 dt1+t2(1(1+2t1+t2)2)=21t=01+t21+4t+6t2+4t3+t4 dt=21t=01+t2(1+t)4 dt

 

Partial fraction decomposition: 1+t2(1+t)4=A1+t+B(1+t)2+C(1+t)3+D(1+t)41+t2=A(1+t)3+B(1+t)2+C(1+t)+D(1)t=1:2=D(2)t=0:1=A+B+C(3)t=1:0=8A+4B+2C|:20=4A+2B+C(4)t=2:3=A+BC(2)+(4):1+3=2BB=1(3)+(4):0+3=3A+3B|:31=A+B|B=11=A+1A=0(2):C=1AB=101C=2 

 

=21t=0(1(1+t)221(1+t)3+21(1+t)4) dt=21t=01(1+t)2 dt41t=01(1+t)3 dt+41t=01(1+t)4 dt=2[11+t]1t=042[1(1+t)2]1t=0+43[1(1+t)3]1t=0=2[12(1)]42[14(1)]+43[18(1)]=2[12+1]2[14+1]+43[18+1]=212234+4378=132+76=6696+76=13696=46=23

 

x=01(x+1+x2)2 dx=23

 

laugh

 May 2, 2017
 #6
avatar+118710 
+2

Thanks Heureka,

I was wondering how this would be done.  

I did think of substituting x=tan(z) but I didn't get very far with it.   

Now I shall take a good look at your answer :))

 May 2, 2017
 #7
avatar+26397 
+2

Thanks Melody

 

laugh

heureka  May 3, 2017

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