Integrals:

Question

Integrals:

hartnn · Answer

$I~\heartsuit ~ Integrals$

epoweritheta · Answer

ya

luigi0210 · Answer

$$\LARGE \int_{1}^{\infty}~\frac{1}{x^2+x}~dx$$

hartnn · Answer

partial fractions

luigi0210 · Answer

$$\LARGE \lim_{t \rightarrow \infty}\int_{1}^{t}(\frac{1}{x}-\frac{1}{x+1})$$?

hartnn · Answer

$\large \dfrac{x+1}{x} =1+  \dfrac{1}{x}$

why do u need to do that ?

hartnn · Answer

no need of taking limits

you can directly use the fact that 
$\dfrac{1}{\infty } =0 $

loser66 · Answer

I am with Lui

hartnn · Answer

ofcourse, you CAN take limits
you'd surely get the correct answer. 
but it will just increase number of steps when you can directly use 1/infinity = 0

loser66 · Answer

@hartnn  Not the same answer.

hartnn · Answer

without limits you won't get log 2 ?

lyrae · Answer

1/infinity is undefined, the limit is not.

loser66 · Answer

@Luigi0210  show us your work, please 
@hartnn  let's wait for Lui to compare the answers and discuss then.

hartnn · Answer

okies! 
use limits, its a full-proof method :)
no shortcuts :P

rational · Answer

technically yes, but if i wanted to figure out the answer in my head, i wont care about taking limit... i wud stickin the infinity straight into the evaluated integral :)

hartnn · Answer

if you're writing an exam, where steps carries marks, then please do use limits.

rational · Answer

but yes limit is the proper definition for these improper integrals

hartnn · Answer

i just assumed you wanted to know the method here...

luigi0210 · Answer

The answer is $ln2$ actually, but I got stuck, maybe I messed up somewhere? 
$$\LARGE \lim_{t \rightarrow \infty}\int_{1}^{t}(\frac{1}{x}-\frac{1}{x+1})$$
$$\LARGE \lim_{t \rightarrow \infty}~(ln|x|-ln|x+1|)|_{1}^{t}$$ 
$$\LARGE \lim_{t \rightarrow \infty}~(ln|t|-ln|1|)-(ln|t+1|-ln|2|)$$

hartnn · Answer

ln A - ln B = ln (A/B)

rational · Answer

$\large \ln | a+b| =  \ln |a| +  \ln | b|$

hartnn · Answer

you sure rational ?

rational · Answer

little bit

rational · Answer

as t and 1 are positive, should not be harmful i think...

hartnn · Answer

did you mean 
$\ln|ab| = \ln|a|+\ln |b|$
?

luigi0210 · Answer

Ohh, those log rules.. 
So does that $\Large \frac{\infty}{\infty}=0$ ?

luigi0210 · Answer

*mean

hartnn · Answer

no

rational · Answer

Oh sh#4 lol yes that one :)

hartnn · Answer

why are u getting infinity/infinity ?

hartnn · Answer

ln 1 = 0

hartnn · Answer

$\ln |t| - \ln |t+1| = -\ln |\dfrac{t+1}{t}| $

loser66 · Answer

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