How do you evaluate the definite integral #int dx / ( x(sqrt(ln(x)))# from #[e, e^81]#?

Answer 1

16

just as an alternative approach,

for #int_e^(e^81)dx/(xsqrtln(x))#

if you recognise the pattern

#d/dx (sqrtln(x) )#
#= 1/2 1/ sqrt (ln( x)) * 1/x#

then the integral is

#int_e^(e^81) d/dx (2 sqrtln(x)) \ dx#
#=2 [ sqrtln(x) ]_e^(e^81)#
#2( [ sqrt(81)] - [ sqrt(1) ] ) = 16#
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Answer 2

#int_e^(e^81)dx/(xsqrtln(x))=16#

We want to find:

#int_e^(e^81)dx/(xsqrtln(x))#
We will want to use substitution. Note that if we let #u=ln(x)#, then #(du)/dx=1/x# and #du=dx/x#, which is currently present in the integral.
Before making these substitutions, note that making the substitutions will necessitate that the boundaries change. To change them, plug the current bounds into #u=ln(x)#.
We see that #e# becomes #u(e)=ln(e)=1# and #e^81# becomes #u(e^81)=ln(e^81)=81#.
Thus, since #color(red)(u=ln(x))# and #color(blue)(du=dx/x)#:
#int_e^(e^81)color(blue)dx/(color(blue)xsqrtcolor(red)ln(x))=int_1^81(du)/sqrtu#

Rewrite the integral:

#int_1^81(du)/sqrtu=int_1^81u^(-1/2)du#

Which can be integrated using the rule:

#int_a^bu^ndu=[u^(n+1)/(n+1)]_a^b" "" ",n!=-1#

So:

#int_1^81u^(-1/2)du=[u^(-1/2+1)/(-1/2+1)]_1^81=[2sqrtu]_1^81=2sqrt81-2sqrt1=16#
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Answer 3

To evaluate the definite integral ∫ dx / (x√(ln(x))) from [e, e^81], we first need to find the antiderivative of the integrand. Then, we can use the Fundamental Theorem of Calculus to evaluate the integral over the given interval.

Let u = ln(x), then du = (1/x) dx. Substituting u = ln(x) and du = (1/x) dx into the integral, we get:

∫ dx / (x√(ln(x))) = ∫ du / (√u)

This is a standard integral, which evaluates to 2√u + C, where C is the constant of integration.

Now, substituting back u = ln(x), we get:

2√(ln(x)) + C

Now, we evaluate this antiderivative at the upper and lower bounds of integration:

At x = e: 2√(ln(e)) = 2√(1) = 2

At x = e^81: 2√(ln(e^81)) = 2√(81) = 2 * 9 = 18

Now, subtract the value at the lower bound from the value at the upper bound to find the definite integral:

Definite Integral = 18 - 2 = 16.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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