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

1. VeritasVosLiberabit

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

2. VeritasVosLiberabit

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

3. Rosh007

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. Rosh007

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

5. SABREEN

is 1/a the derivative of x/a

6. SABREEN

is tht why its there?

7. VeritasVosLiberabit

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

8. VeritasVosLiberabit

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

9. SABREEN

thnk you id like to see why

10. VeritasVosLiberabit

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

11. VeritasVosLiberabit

in this case z is a scalar.

12. VeritasVosLiberabit

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

13. SABREEN

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

14. VeritasVosLiberabit

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. Rosh007

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. SABREEN

it is 1/1+x^2

17. Rosh007

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

18. SABREEN

i dunno

19. Rosh007

20. Rosh007

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

21. SABREEN

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

22. Rosh007

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

23. malevolence19

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

24. SABREEN

a/(x2+a^2)

25. SABREEN

yea shure if its not too complicated

26. Rosh007

27. Rosh007

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

28. malevolence19

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. Rosh007

now itegrat on both sides.. you got answer

30. Rosh007

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

31. SABREEN

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

32. SABREEN

thanks=)

33. Rosh007

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

34. malevolence19

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

35. SABREEN

loll nahh i got this bro=)

36. Rosh007

@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. malevolence19

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)