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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.

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.

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

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

now itegrat on both sides.. you got 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=)