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

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

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\[\int\limits_{}^{}\frac{ z }{ x ^{2}+z ^{2} }dx\]
this is a formula you use to compute the integral of a function with this structure.
You will be able to relate your question simply to the standard integral \[\large \frac{1 }{ a^2+x^2 } =\frac{1 }{a } \times \arctan (\frac{x}{ a })\]

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Other answers:

if Left hand side is the function that has to integrate with respect to x
is 1/a the derivative of x/a
is tht why its there?
1/a is part of the standard formula. I'll work it out for you.
\[z \int\limits_{}^{}\frac{ 1 }{ x ^{2}+z ^{2} }dx \]
thnk you id like to see why
\[z*\frac{ 1 }{ z }\arctan(\frac{ x }{ z })\rightarrow \arctan(\frac{ x }{ z })\]
in this case z is a scalar.
so typically z would be represented as a number. I think they are using zahlen which denotes integer values.
why is arc tan (x/z) and where does the 1/z come from??
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.
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 ??
it is 1/1+x^2
well,good.. then what will be derivative of arctan(x/a)
i dunno
ok.. instead of x it is x/a.. substitute that value in your answer above
and rember x/a is another function so you need to apply the chain rule of differentiation
1/(1+(x/a)^2)*(1/a)
Yup........... You got it.. simplify it
I mean, I can do the integral if you want it in all the gruesome detail...
a/(x2+a^2)
yea shure if its not too complicated
well you are right ... now compare with your question......you got your answer
that is you got \[\frac{ d }{ dx }(\arctan(x/a) = \frac{ a }{ x^2+a^2 }\]
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) :/
now itegrat on both sides.. you got answer
Remember Integration is the opposite of differentiation @SABREEN .. now what will be tha answer?
so than the integral of a/(x^2+a^2) is arctan (x/a)
thanks=)
yes.. Welcome.. Now should I explain how 1/a comes ???? ;)
Next one in line: \[\int\limits \tanh^{-1}(ax)dx\]
loll nahh i got this bro=)
@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.. :)
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)

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