intergration (1 Viewer)

P.T.F.E

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

study-freak

Bored of
Look at my attachment...

let u=1/x ....

tommykins

i am number -e^i*pi
$\bg_white \int_{0.5}^{2} \frac{lnx}{1+x^2}\\ u = lnx \rightarrow u' = \frac{1}{x}\\ v' = \frac{1}{1+x^2} \rightarrow v = \tan^{-1}x\\ I = lnx.tan^{-1}x - \int \frac{tan^{-1}x}{1+x^2}\\ = lnx.tan^{-1}x - \frac{(tan^{-1}x)^2}{2}$

then keep going putting in the limits

Iruka

Member
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

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

Check my working...

lolokay

Active Member
yep, that working's correct

P.T.F.E

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

youngminii

Banned
lolokay is a genius.

P.T.F.E

Member
but whats the area!!!!!

0

MC Squidge

BOS' Apex Predator
genius substitution

study-freak

Bored of
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.

^
no dice

P.T.F.E

Member
can somebodydeduce the value of the intergrel

lolokay

Active Member
can somebodydeduce the value of the intergrel

P.T.F.E

Member
woops my bad i missed it ha thanks all