## anonymous 3 years ago how do we integrate z/(x^2+z^2) with respect to x the answer is arctan(x/z)

1. anonymous

$\int\limits_{}^{}\frac{ z }{ x ^{2}+z ^{2} }dx$

2. anonymous

this is a formula you use to compute the integral of a function with this structure.

3. anonymous

You will be able to relate your question simply to the standard integral $\large \frac{1 }{ a^2+x^2 } =\frac{1 }{a } \times \arctan (\frac{x}{ a })$

4. anonymous

if Left hand side is the function that has to integrate with respect to x

5. anonymous

is 1/a the derivative of x/a

6. anonymous

is tht why its there?

7. anonymous

1/a is part of the standard formula. I'll work it out for you.

8. anonymous

$z \int\limits_{}^{}\frac{ 1 }{ x ^{2}+z ^{2} }dx$

9. anonymous

thnk you id like to see why

10. anonymous

$z*\frac{ 1 }{ z }\arctan(\frac{ x }{ z })\rightarrow \arctan(\frac{ x }{ z })$

11. anonymous

in this case z is a scalar.

12. anonymous

so typically z would be represented as a number. I think they are using zahlen which denotes integer values.

13. anonymous

why is arc tan (x/z) and where does the 1/z come from??

14. anonymous

1/z is always part of the equation. if you want me to do a proof, I'm afraid I don't have enough time for that but maybe someone else would.

15. anonymous

the formula for most of the integral is coming fromits derivative remember derivation is the opposite of integration and wise versa. can you say what is th derivative of f(x) = arctan x ??

16. anonymous

it is 1/1+x^2

17. anonymous

well,good.. then what will be derivative of arctan(x/a)

18. anonymous

i dunno

19. anonymous

20. anonymous

and rember x/a is another function so you need to apply the chain rule of differentiation

21. anonymous

1/(1+(x/a)^2)*(1/a)

22. anonymous

Yup........... You got it.. simplify it

23. anonymous

I mean, I can do the integral if you want it in all the gruesome detail...

24. anonymous

a/(x2+a^2)

25. anonymous

yea shure if its not too complicated

26. anonymous

27. anonymous

that is you got $\frac{ d }{ dx }(\arctan(x/a) = \frac{ a }{ x^2+a^2 }$

28. anonymous

I mean, there isn't much too it though. The formula is basically where it comes from... Now that I think about it, I don't know how much you can actually write. I was running it by wolfram and all they do is a variable substitution and then use that int (1/u^2+1)=arctan(u) :/

29. anonymous

now itegrat on both sides.. you got answer

30. anonymous

Remember Integration is the opposite of differentiation @SABREEN .. now what will be tha answer?

31. anonymous

so than the integral of a/(x^2+a^2) is arctan (x/a)

32. anonymous

thanks=)

33. anonymous

yes.. Welcome.. Now should I explain how 1/a comes ???? ;)

34. anonymous

Next one in line: $\int\limits \tanh^{-1}(ax)dx$

35. anonymous

loll nahh i got this bro=)

36. anonymous

@SABREEN ....Good... Welcome and enjoy.... @malevolence19 .......you could solve by following the method i described here...only difference is x = ax and remember ax is a fuction.. :)

37. anonymous

I mean, I know how to to do it lol. It's a integration by parts problem once you look up the derivative to tanh^(-1)(x)