1
$\begingroup$

How can we solve $\displaystyle\int\frac{1}{(x^2+1)^2}dx$ for Riemanian integration and Lebesque integration mode?

$\endgroup$

closed as off-topic by Nosrati, Martín Vacas Vignolo, Archer, José Carlos Santos, Andrew Dec 30 '18 at 17:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Martín Vacas Vignolo, Archer, José Carlos Santos, Andrew
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ This is an indefinite integral! $\endgroup$ – Mhenni Benghorbal Dec 21 '13 at 23:48
  • $\begingroup$ Are there any bounds you're forgetting, or is this actually indefinite? $\endgroup$ – cygorx Dec 22 '13 at 0:32
3
$\begingroup$

Hint: Use the change of variables $x=\tan t$ to get

$$ \int \cos^2 t\, dt .$$

I think you can finish it.

Note: The following identity is useful

$$ \cos(2x)=2\cos^2 x - 1 .$$

$\endgroup$
  • 1
    $\begingroup$ @Maryam: Have you studied Riemann integration and Lebesgue integration? $\endgroup$ – Mhenni Benghorbal Dec 22 '13 at 0:07
0
$\begingroup$

A method worth looking at (now I do not know its ramifications for Riemannian Integration or Lebesgue Integration, someone else will need to explore that) is good old partial fractions from Calc 2:

$$(1 + x^2) = (1 + ix) (1 - ix)$$

$$ \int{\frac{1}{(1 + x^2)^2}} = \int{\frac{a_1}{1 + ix}} + \int{\frac{a_2}{1 + ix}} + \int{\frac{a_3}{1 - ix}} + \int{\frac{a_4}{1 - ix}}$$

Such that:

$$a_i = u_ix + v_i$$

The challenge then is to simplify the expression you get (which may involve a variety of complex logarithms) into something clean involving, preferably, inverse trig functions.

But of course, if Lebesgue and Riemman integration is only defined over $R$ then this is of no use to you.

$\endgroup$
0
$\begingroup$

Here is a partial fraction expansion that may be useful

$$\frac{1}{(1+x^2)^2} = \frac{1}{2} \frac{1}{1+x^2} - \frac{1}{2} \frac{x^2-1}{(1+x^2)^2}$$

Riemannian integral is easily obtained as $$\frac{1}{2} atan(x) + \frac{1}{2} \frac{x}{1+x^2}$$

Hope this helps BTW: I followed the usual approach to partial fractions and then regrouped the terms into something nice.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.