# 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