nubeer
integrate (1/(x^2+8x +25 )^0.5)



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Andras
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This is a hard one
x^2+8x+25=(x+4)^2+9
so use sub u=x+4 du=dx
New integral:
1/(u^2+9)^0.5

nubeer
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hmmm ya i also reached at this point but what after this?

Andras
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thinking of the next substitution, maybe some trig function

nubeer
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hmm yes i think that would be tan but how and why

Andras
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I know! Recall that tan^2(x)+1=sec^2(x)

Andras
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so we need to make 9tan^2(x) from u to get 9(tan^2(x)+1)

Andras
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let u=3tan(z)
du=3sec^2(z) dz

Andras
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So the integral will be:
3sec^2(z)/3sec(z)=sec(z)

Andras
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What the integral of secz? log(tanz+secz) (just looked it up)

Andras
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we have u=3tanz expressing z=tan^1(u/3)

nubeer
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thanks man.. well pretty much hit the answer but can u explain this 1 step let u=3tan(z) du=3sec^2(z) dz.. why we suppose 3 tan(z) why not just tan

Andras
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because if it is 3tan than (3tan)^2=9tan^2(x) so we can factor out 9 to get 9(tan^2(x)+1)

nubeer
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ohh so just that was the reason to put 3?

Andras
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yes

Andras
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Hope it helped, now you just need to plug back the z and u to get x

FoolForMath
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do you know about hyperbolic functions ?

FoolForMath
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sinhx ?

FoolForMath
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and this is a elementary problem no need to think hard on this one

Andras
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e^x+e^x/2
how would that help?

Zarkon
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\[\frac{d}{dx}\operatorname{arsinh}(x) =\frac{1}{\sqrt{x^{2}+1}}\]