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Could someone list out the steps, I managed to find the equation in terms of x. But im having trouble integrating sqrt (x^2+4x)

 May 3, 2017
 #1
avatar+118710 
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This has already been answered here

 

https://web2.0calc.com/questions/one-more

 May 3, 2017
 #2
avatar+26398 
+2

Need help with this continuous fraction integral.

integral \limits_{x=0}^1  \frac{x\pm\sqrt{x^2+4x} }{2}  \ dx

 

Continuous Fraction.

sum=x+xx+xx+xx+sum=x+xsumsumxsum=xsum2xsum=xsum2x=xsumsum2xsumx=0sum=x±x24(x)2sum=x±x2+4x2

 

1x=0x+xx+xx+xx+ dx=1x=0sum dx=1x=0x±x2+4x2 dx=121x=0x dx±121x=0x2+4x dx=12[x22]1x=0±121x=0(x+2)24 dx=14±121x=0(x+2)24 dxsubstitute: u=x+2du=dx new limits: x=0:u=0+2u=2x=1:u=1+2u=3 =14±123u=2u24 dusubstitute: u=2cosh(z)z=arcosh(u2)du=2sinh(z) dz new limits: u=2:z=arcosh(22)z=arcosh(1)=0u=3:z=arcosh(32)z=arcosh(1.5) =14±12arcosh(1.5)z=04cosh(z)242sinh(z) dz=14±arcosh(1.5)z=04cosh(z)24sinh(z) dz cosh2(z)sinh2(z)=1cosh2(z)1=sinh2(z)|44cosh2(z)4=4sinh2(z) =14±arcosh(1.5)z=04sinh2(z)sinh(z) dz=14±arcosh(1.5)z=02sinh(z)sinh(z) dz=14±arcosh(1.5)z=02sinh2(z) dz

 cosh(2z)=cosh2(z)+sinh2(z)cosh(2z)=1+sinh2(z)+sinh2(z)cosh(2z)=1+2sinh2(z)2sinh2(z)=cosh(2z)1 =14±arcosh(1.5)z=0(cosh(2z)1) dz=14±(arcosh(1.5)z=0cosh(2z) dzarcosh(1.5)z=0 dz)=14±([12sinh(2z)]arcosh(1.5)z=0[z]arcosh(1.5)z=0) sinh(2z)=2sinh(z)cosh(z) =14±([122sinh(z)cosh(z)]arcosh(1.5)z=0arcosh(1.5))=14±([sinh(z)cosh(z)]arcosh(1.5)z=0arcosh(1.5))=14±[sinh(arcosh(1.5))1.5arcosh(1.5)] cosh2(x)sinh2(x)=1sinh2(x)=cosh2(x)1sinh(x)=cosh2(x)1  sinh(arcosh(x))=cosh2(arcosh(x))1=x21sinh(arcosh(1.5))=1.521=1.25=52 =14±[5232arcosh(1.5)]=14±[354arcosh(1.5)]=14±(1.67705098312484240.9624236501192069)=14±0.7146273330056354

 

=14+0.7146273330056354or=140.7146273330056354=0.9646273330056354or=0.4646273330056354

 

laugh

 May 4, 2017
edited by heureka  May 4, 2017

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