How to solve the indefinite integral of the function y=/x*square root[(ln x)^2-1]?



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giorgiana1976's profile pic

Posted on (Answer #1)

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

hala718's profile pic

Posted on (Answer #2)

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

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