help with definite integral?!?

. 0

∫ x/(x^2-4) dx = (1/2) [ ln( (0)^2 - 4 ) - ln( (-√3)^2 - 4 ) ] = (1/2) ln(4) ~ .6931

-√3

I get

1/2 ln|x^2 - 4|

evaluate

1/2 ln|0^2 - 4| - 1/2 ln|3 - 4|

= (1/2)(ln(4)) - (1/2)ln(1)

= ln(sqrt(4)) - 0

= ln(2)

the function can be seen to be positive over all the range, so the answer should certainly be a positive value

How dare you call my solution incorrect, I checked and it is correct.

Integrate the function and then plug in the limits:

∫ [0, -√3] x / (x² - 4) dx = ∫ [0, -√3] 2x / (x² - 4) dx / 2

∫ [0, -√3] x / (x² - 4) dx = [ln|x² - 4|][0, -√3] / 2

∫ [0, -√3] x / (x² - 4) dx = ln4 / 2

∫ [0, -√3] x / (x² - 4) dx = ln2² / 2

∫ [0, -√3] x / (x² - 4) dx = 2ln2 / 2

∫ [0, -√3] x / (x² - 4) dx = ln2

Opinions don't really have much place here. The answer is ½ln(2) which may also be expressed as

ln(√(2)).

Use the substitution u = x² - 4, du = 2x dx. The integral becomes

-4

∫ ½ du/u

-1

∫x/(x^2-4)dx (from -sqrt3 to 0) = ?

Let, x^2-4 = u, then x.dx = du/2

∫x/(x^2-4)dx (from -sqrt3 to 0) = ∫du/2(u)

∫x/(x^2-4)dx (from -sqrt3 to 0) = (1/2)ln(u) = (1/2)ln(x^2-4)

∫x/(x^2-4)dx (from -sqrt3 to 0) = (1/2)[ln(-3-4) - ln(-4)]

∫x/(x^2-4)dx (from -sqrt3 to 0) = (1/2)ln[-7/-4]

∫x/(x^2-4)dx (from -sqrt3 to 0) = (1/2)ln(7/4) = 0.5596/2

∫x/(x^2-4)dx (from -sqrt3 to 0) = 0.2798 >==========< ANSWER

x/(x^2-4) =x/(x+2)(x-2) as x^2 -4 factorises to (x+2)(x-2) using partial fraction we get

x/(x^2-4)= 1/2(x+2)+1/2(x-2) integrating this gives 1/2 ln|(x+2)|+1/2ln| (x-2)|

this simplifies to

1/2ln (x+2)(x-2) =1/2ln|(x^2-4)|

when x= sqrt3 this gives ln1 which is zero

when x=0 this gives 1/2ln4 =ln2

so answer is 0-ln2=-ln2

i'm sorry yet Gianlino's evidence isn't valid. x^2 and sqrt(x) are inverse of one yet another, although their integrals from 0 to a million are respectively a million/3 and a pair of/3 extra in many situations, the graphs of purposes inverse of one yet another would be symmetrical wrt the 1st bisector. for this reason the integrals, measuring the section between those graphs and the x axis and the strains x=0 and x=a million, will in many situations no longer be equivalent. they are in a position to be so in easy terms in specific situations. i think of that's extremely user-friendly. permit me have a attempt. Dina, your substitution is only approximately the final one, seem With u^3 = a million-x^7 J = Int(0,a million)(a million-x^7)^(a million/3) dx transforms as follows: (a million-x^7)^(a million/3) --> u dx --> d(a million-u^3)^(a million/7) So J = Int(a million,0)ud(a million-u^3)^(a million/7) = u(a million-u^3)^(a million/7)|_1^0 - Int_1^0 (a million-u^3)^(a million/7)du = 0 + your 2nd imperative so that they are equivalent ! you should apply the comparable trick for any (m,n) helpful integers quite than (3,7) Edited: ok, my argument approximately gianlino's exchange into incorrect through fact i did no longer pay interest to his specifying reducing purposes. And the exchange of variables I proposed is strictly changing functionality and variable (subsequently utilising the inverse functionality) and is an illustration of gianlino's declare. i choose some cafeine

Ans: 1/2 . (log 2 + 1/log2) . [1/log(root 3) - 1]