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How do you evaluate the integral #int 1/(x(1+(lnx)^2)#?

Answer 1

Christopher Allers

#int1/(x(1+(lnx)^2))dx=arctan(lnx)+C#

Let #u=lnx#. This implies that #du=1/xdx#.

#int1/(x(1+(lnx)^2))dx=int1/(1+(lnx)^2)(1/xdx)=int1/(1+u^2)du#

This is the arctangent integral:

#=arctan(u)+C=arctan(lnx)+C#

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Answer 2

Ezekiel Alexander

To evaluate the integral (\int \frac{1}{x(1+(\ln x)^2)}), we can use the substitution method. Let (u = \ln x), then (du = \frac{1}{x} dx). Substituting these into the integral gives:

[ \int \frac{1}{x(1+(\ln x)^2)} , dx = \int \frac{1}{1+u^2} , du ]

Now, this integral is in a form that we can easily integrate using the arctangent function:

[ \int \frac{1}{1+u^2} , du = \arctan(u) + C ]

Finally, substituting back (u = \ln x) and simplifying gives the result:

[ \int \frac{1}{x(1+(\ln x)^2)} , dx = \arctan(\ln x) + C ]

where (C) is the constant of integration.

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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.