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

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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
hmmm ya i also reached at this point but what after this?
thinking of the next substitution, maybe some trig function

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Other answers:

hmm yes i think that would be tan but how and why
I know! Recall that tan^2(x)+1=sec^2(x)
so we need to make 9tan^2(x) from u to get 9(tan^2(x)+1)
let u=3tan(z) du=3sec^2(z) dz
So the integral will be: 3sec^2(z)/3sec(z)=sec(z)
What the integral of secz? log(tanz+secz) (just looked it up)
we have u=3tanz expressing z=tan^-1(u/3)
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
because if it is 3tan than (3tan)^2=9tan^2(x) so we can factor out 9 to get 9(tan^2(x)+1)
ohh so just that was the reason to put 3?
Hope it helped, now you just need to plug back the z and u to get x
do you know about hyperbolic functions ?
sinhx ?
and this is a elementary problem no need to think hard on this one
e^x+e^-x/2 how would that help?
\[\frac{d}{dx}\operatorname{arsinh}(x) =\frac{1}{\sqrt{x^{2}+1}}\]

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