Monkey Raptor

We want to find the solution for the indefinite integral from:

1. $∫ (\ln x)\ dx$

2. $∫ 1/(\ln x)\ dx$

3. $∫ 1/(x.\ln x)\ dx$

4. $∫ 1/(x^{2}.\ln x)\ dx$

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$∫ 1/x\ dx = \ln |x| +\ C$

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• Integral by parts

  $∫ u.dv = u.v - ∫ v.du$

• Substitution - of $1/u$

  $∫ 1/{u}\ du = \ln u +\ C$

• Using (table look-up, or probably remembering some) integral table

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1^st problem

$∫ (\ln x)\ dx$

The solution can be achieved using integral by parts. So then, u = ln x dv = dx du/dx [derivative of u] = 1/x switch the (dx) to right side ► du = (1/x) dx v [integral of dv ► ∫ 1 dx] = x Substitute those variables: ∫ u.dv = u.v - ∫ v.du ∫ (ln x) dx = (ln x).x - ∫ x.(1/x) dx = x.(ln x) - ∫ 1 dx = x.(ln x) - x + C So, ∫ (ln x) dx = x.(ln x) - x + C Or, ∫ (ln x) dx = x(ln x - 1) + C This solution is already included in standard integral table.

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2^nd problem

$∫ 1/(\ln x)\ dx$

This is a special logarithmic integral.
So the solution would be (using integral table):

Or (using jqMath — great with Firefox or other browser which supports MathML)

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3^rd problem

$∫ 1/{x.(\ln x)}\ dx$

This can be achieved using substitution. So, u = ln x du/dx = 1/x switch (dx) to right side ► du = (1/x) dx Substitute those variables: ∫ (1/u)du = ln u + C ∫ 1/[x.(ln x)] dx = ln [ln x] + C This solution also already exists in integral table.

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Last problem

$∫ 1/{{x^2}.(\ln x)}\ dx$

Let's take a look again at the integral table. It magically reveals this form:

Or

You can substitute the: m with (-2): for x and n with (-1): for ln x ----------------------- Then that's the solution

Note: you don't have to "solve" the second integral, it will be endless. BUT, if you have so much idle time, you may scribble some more.

You can go to Wolfram|Alpha to find out the graphs of those problems: over there

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So, in conclusion, finding the right method and "remembering" integral table are important to solve those kind of problems.