integration question (2 Viewers)

[who_loves_maths]

Originally Posted by Slide_Rule
u=sqrt(x^2-4)
du=1/2.x/sqrt(x^2-4)

where did the 1/2 come from? if u=sqrt(x^2-4), then du = xdx/sqrt(x^2-4)
so I = 2Int[(ln(u))du]

Originally Posted by Slide_Rule
u=sqrt(x^2-4)
du=1/2.x/sqrt(x^2-4)

Int(1/2ln(u^2))du
=Int(ln(u))du

even if du =1/2.x/sqrt(x^2-4) , then the following integral should be Int[2ln(u^2)du] isn't it?

 

[polythenepam]

hmm yeeh
i asked my teacher
he said the question was missing an x..
it turns out that it comes from a retyped catholic trial so somebody made a typo

 

[wori]

substitution+integrate by parts

2*SUM{ln[(x^2-4)^0.5]/[(x^2-4)]}dx=2*SUM f(x)*f(x)' dx
where f(x)=ln[(x^2-4)^0.5]

then use integrating by parts
let u=ln[(x^2-4)^0.5]