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\[y = \int\limits_{} \tan^{-1}(t) dt\]
is this the equation?

yes but it is a definite integral 1/x to 4x

does that make sense so far?

i wonder if we can decompose this......

I dont see any easy ways of doing this :)

i think i'm just making it more difficult than it needs to be in some way

do you understand piecewise functions?

is there an answer that you can chaeck against?

i know if piecewise functions; but havenet come across a way to integrate them

mmk and there isn't an answer to check :/ thanks for trying though!

.-. I don't think that's right amistre though I didn't try it myself lol >_<

wh are you letting tan^-1(t) = u? ._.

becasue there is no function that derives down to a tan^-1

it is equivalent to:
[S] u sec^2(u) du tho.... and thats doable somehow

lol I got you amistre :)

int by parts is still a mystery to me for some functions :)

O_O what kind of table is THAT LOL

.-. why so complicated? LOL

you can do it my way :) much easier ^_^

>_< lol, I've learned something new today! thank you :)

:) all I need is an integral for tan(u) and I got it solved lol

the table of integration says: tan(x) ints to ln(sec(x)) cool :)

u tan(u) - ln(sec(u))

hmm:
\[\tan x = \frac{sinx}{cosx}\]
let u = cos x and du = -sinx and solve lol

we re-stitute for u and get:
tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))

.-.

well...... ln(sec(tan^-1(t))) lol got excited ;)

not convinced with the sec u amistre :(

can always try to derive it back ;)

tan x = sinx/cosx
integrating that will give you ln|cosx|

>_

\[\tan ^{-1}t=u, tanu = t\]
\[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]

._.

:) you too

:)

\[\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u\]

integrating by parts
\[\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]

\[utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})\]

if I did it right; mine derives back to tan^-1(t)....

\[utan ^{-1}u - 2 \ln (1+u ^{2})\]

i used integration by parts method, not substitution

so the result?

\[\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}\]

n this is what u wanted?

i guess i should terminate at Tan^(-1)t?

since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with..
tan^-1(t)

...since t^2 + 1.....typoed it :)

hmmmm,right :)

did u check mine?is it wrong?

yup...hats off to ur patience :P