Question

how do we integrate z/(x^2+z^2) with respect to x
the answer is arctan(x/z)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

is 1/a the derivative of x/a

Answer 6

is tht why its there?

Answer 7

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

Answer 8

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

Answer 9

thnk you id like to see why

Answer 10

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

Answer 11

in this case z is a scalar.

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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 ??

Answer 16

it is 1/1+x^2

Answer 17

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

Answer 18

i dunno

Answer 19

ok.. instead of x it is x/a.. substitute that value in your answer above

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

a/(x2+a^2)

Answer 25

yea shure if its not too complicated

Answer 26

well you are right ... now compare with your question......you got your answer

Answer 27

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

Answer 28

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) :/

Answer 29

now itegrat on both sides.. you got answer

Answer 30

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

Answer 31

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

Answer 32

thanks=)

Answer 33

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

Answer 34

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

Answer 35

loll nahh i got this bro=)

Answer 36

@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.. :)

Answer 37

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)