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i did this with u substituion it did not work
i am getting something like ln2/2

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Other answers:

It looks to me more an atan(2x)
how
i took u=1+4t^2
du/8t=dt
The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)
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So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)
use the equation button
The result is not ln(u)
ok i should remember that formula right?
I suggest you use y=2t
Yes, it's on the first page of standard integrals. Check it out.
ok
\( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)
@mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c\]
Right, so a=1/2.
what is x then
\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]
x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)
and a=1/2
can i do \[\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}\]
and t=1/2
t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.

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