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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

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)

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?

yes

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?

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