intergration (1 Viewer)

P.T.F.E · May 24, 2009

intergral of lnx/1+x^2 from .5 to 2

study-freak · May 24, 2009

Look at my attachment...

lolokay · May 24, 2009

let u=1/x ....

tommykins · May 24, 2009

$\int_{0.5}^{2} \frac{lnx}{1+x^2}\\ u = lnx \rightarrow u' = \frac{1}{x}\\ v' = \frac{1}{1+x^2} \rightarrow v = \tan^{-1}x\\ I = lnx.tan^{-1}x - \int \frac{tan^{-1}x}{1+x^2}\\ = lnx.tan^{-1}x - \frac{(tan^{-1}x)^2}{2}$

then keep going putting in the limits

Iruka · May 24, 2009

I think you've made a mistake in the third line.

lolokay's method is far more interesting - just make sure that you substitute for the limits of integration as well...

Trebla · May 24, 2009

Elaborating on what lolokay suggested:
$\int_{0.5}^2 \dfrac{\ln x}{1+x^2}\,dx\\\\x=\dfrac{1}{u}\Rightarrow dx=-\dfrac{1}{u^2}du\\\\\int_{0.5}^2 \dfrac{\ln x}{1+x^2}\,dx=\int_{2}^{0.5} \dfrac{-\ln \frac{1}{u}}{1+\frac{1}{u^2}}\dfrac{1}{u^2}\,du\\\\=\int_{2}^{0.5} \dfrac{\ln u}{1+u^2}\,du\\\\= - \int_{0.5}^{2} \dfrac{\ln x}{1+x^2}\,dx\,\,\,\,($since $u$ is a dummy variable$)\\\\\therefore 2\displaystyle\int_{0.5}^2 \dfrac{\ln x}{1+x^2}\,dx=0\\\\\Rightarrow \displaystyle\int_{0.5}^2 \dfrac{\ln x}{1+x^2}\,dx=0$

Check my working...

lolokay · May 24, 2009

yep, that working's correct

P.T.F.E · May 24, 2009

so whts the the answer. we were asked to prove the 'u" bit but what is the answer

youngminii · May 24, 2009

lolokay is a genius.

gurmies · May 24, 2009

youngminii said:
lolokay is a genius.

Correct.

P.T.F.E · May 24, 2009

but whats the area!!!!!

Timothy.Siu · May 24, 2009

MC Squidge · May 24, 2009

genius substitution

riseek · May 25, 2009

lolokay said:
let u=1/x ....

genius

maths-bbqz · May 25, 2009

study-freak said:
Look at my attachment...

Wow ur so cool.
faggot

study-freak · May 25, 2009

maths-bbqz said:
Wow ur so cool.
faggot

You don't have to introduce yourself in this forum. You are allowed to be anonymous and unknown to others.

MC Squidge · May 25, 2009

^
no dice

P.T.F.E · May 25, 2009

can somebodydeduce the value of the intergrel

lolokay · May 25, 2009

P.T.F.E said:
can somebodydeduce the value of the intergrel

..has already been posted (twice) in the thread

P.T.F.E · May 25, 2009

woops my bad i missed it ha thanks all

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