## nubeer 4 years ago integrate (1/(x^2+8x +25 )^0.5)

1. Andras

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

2. nubeer

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

3. Andras

thinking of the next substitution, maybe some trig function

4. nubeer

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

5. Andras

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

6. Andras

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

7. Andras

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

8. Andras

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

9. Andras

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

10. Andras

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

11. nubeer

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

12. Andras

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

13. nubeer

ohh so just that was the reason to put 3?

14. Andras

yes

15. Andras

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

16. FoolForMath

do you know about hyperbolic functions ?

17. FoolForMath

sinhx ?

18. FoolForMath

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

19. Andras

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

20. Zarkon

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