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How to solve the indefinite integral of the function y=/x*square root[(ln x)^2-1]?
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To solve the indefinite integral, we'll replace ln x by t:
ln x = t
We'll differentiate both sides:
dx/x = dt
We'll re-write the integral in the new variable t:
Int dx/x*sqrt[(lnx)^2 - 1] = Int dt/sqrt(t^2-1)
Int dt/sqrt(t^2-1) = ln|t + sqrt(t^2-1)| + C
The indefinite integral is: Int dx/x*sqrt[(lnx)^2 - 1] = ln|ln x + sqrt[(lnx)^2 - 1]| + C
Posted by giorgiana1976 on April 26, 2011 at 2:24 AM (Answer #1)
High School Teacher
y= 1/xsqrt(lnx^2 -1)
Let u= ln x ==> du = 1/x dx ==> dx = x du
==> y= 1/x* sqrt(u^2-1)
==> Int y= Int 1/x *sqrt(u^2-1) * x du
==> Int y= Int 1/ sqrt(u^2-1) du
= ln l u+ sqrt(u^2-1) l + C
Now w will substitute with u= ln x
==> Int y= ln l lnx +sqrt (lnx)^2 -1) l + C
Posted by hala718 on April 26, 2011 at 2:33 AM (Answer #2)
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