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ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0i did this with u substituion it did not work

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0i am getting something like ln2/2

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1It looks to me more an atan(2x)

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358025284718:dw

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0use the equation button

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1The result is not ln(u)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0ok i should remember that formula right?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1I suggest you use y=2t

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, it's on the first page of standard integrals. Check it out.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1\( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0@mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{1} \frac{ x }{ a }+c\]

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{1} \frac{ x }{ a}+c\]

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0can i do \[\frac{ 1 }{ 1 }\tan^{1} \frac{ 2t }{ 1}\]

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1t 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 lefthand side is exactly your question. All you need to do after that is to evaluate the righthand side.
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