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Integration....help
- Thread starter Ton T.
- Start date
Note: the bounds are from 0 to +infinty, not from +infinity to 0.
Like you said, this problem is similar to the one posted by Mr. Ne Ed Help, and coincidentally you can use the same method.
Perform integration by parts. Let
. . . . u=tan^(-1)x
and
. . . . dv=x/(1+x^2)^2 dx.
Like you said, this problem is similar to the one posted by Mr. Ne Ed Help, and coincidentally you can use the same method.
Perform integration by parts. Let
. . . . u=tan^(-1)x
and
. . . . dv=x/(1+x^2)^2 dx.
Hello, Ton T.!
It is similar to Mr. Ne Ed help's problem, but <u>much</u> more complicated.
. . I found a way . . . maybe someone has a better approach?
. . . . . . . . . . . . . . . . . . . . . . . x dx
Let: u = tan<sup>-1</sup>x . . . . . dv = -----------
. . . . . . . . . . . . . . . . . . . . . .(1 + x<sup>2</sup>)<sup>2</sup>
. . . . . . . . . . . dx . . . . . . . . . . . . 1
Then: du = -------- . . . . v = - -----------
. . . . . . . . . 1 + x<sup>2</sup> . . . . . . . . 2(1 + x<sup>2</sup>)
. . . . . . . . . . . .tan<sup>-1</sup>x . . . . 1 . . . . . . . .dx
We have: . - ----------- . + .--- [INT] ------------
. . . . . . . . . . .2(1 + x<sup>2</sup>) . . . 2 . . . . . (1 + x<sup>2</sup>)<sup>2</sup>
This new integral can be solved with a Trig. Substitution: .x = tanθ
[Edit: Too fast for me again, Matt!]
It is similar to Mr. Ne Ed help's problem, but <u>much</u> more complicated.
. . I found a way . . . maybe someone has a better approach?
We do it "by parts", of course.int (0 to +Infinity) [x tan<sup>-1</sup>x] / [(1 + x<sup>2</sup>)<sup>2</sup>] dx
.
. . . . . . . . . . . . . . . . . . . . . . . x dx
Let: u = tan<sup>-1</sup>x . . . . . dv = -----------
. . . . . . . . . . . . . . . . . . . . . .(1 + x<sup>2</sup>)<sup>2</sup>
. . . . . . . . . . . dx . . . . . . . . . . . . 1
Then: du = -------- . . . . v = - -----------
. . . . . . . . . 1 + x<sup>2</sup> . . . . . . . . 2(1 + x<sup>2</sup>)
. . . . . . . . . . . .tan<sup>-1</sup>x . . . . 1 . . . . . . . .dx
We have: . - ----------- . + .--- [INT] ------------
. . . . . . . . . . .2(1 + x<sup>2</sup>) . . . 2 . . . . . (1 + x<sup>2</sup>)<sup>2</sup>
This new integral can be solved with a Trig. Substitution: .x = tanθ
[Edit: Too fast for me again, Matt!]
Thank you for pointing me the way. I feel really stupid to ask you this, but I am still not see the end of this problem.
. . . . . . . . . . . .tan-1x . . . . 1 . . . . . . . .dx
We have: . - ----------- . + .--- [INT] ------------
. . . . . . . . . . .2(1 + x2) . . . 2 . . . . . (1 + x2)2
Substitute U = tan-1x so du = 1 / (1+x^2)dx still how can I integrate with (1+x^2)2? Sorry I am really bad in math.
. . . . . . . . . . . .tan-1x . . . . 1 . . . . . . . .dx
We have: . - ----------- . + .--- [INT] ------------
. . . . . . . . . . .2(1 + x2) . . . 2 . . . . . (1 + x2)2
Substitute U = tan-1x so du = 1 / (1+x^2)dx still how can I integrate with (1+x^2)2? Sorry I am really bad in math.