Integral (ln x), 1/(ln x), 1/(x.ln x), and 1/(x².ln x)

Hi. 👋

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. $∫ 1/(x^{2}.\ln x)\ dx$

We need to keep in mind that:

$∫ 1/x\ dx = \ln |x| +\ C$

Well, that's a definition.

Right then.

To solve each of those, we may employ one of these methods:

• Integral by parts ⬇️

  $∫ u.dv = u.v - ∫ v.du$

• Substitution of $1/u$ ⬇️

  $∫ 1/{u}\ du = \ln u +\ C$

• Using integral table look-up, or probably remembering bits from integral table.

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First Problem

$∫ (\ln x)\ dx$

The solution can be achieved using integral by parts. 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 Therefore — ∫ (ln x) dx = x.(ln x) - x + C Or — ∫ (ln x) dx = x(ln x - 1) + C This solution is in standard integral table.

$∫ (\ln x)\ dx = x(\ln x - 1) + C$

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Second Problem

$∫ 1/(\ln x)\ dx$

This is a special logarithmic integral.

So the solution would be — using integral table:

Or —

$$∫{{dx}/{\ln x}}={\ln |\ln x|} + \ln x +Σ↙{k=2}↖∞{(\ln x)^k/{k.k!}}$$

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Third Problem

$∫ 1/{x.(\ln x)}\ dx$

This can be achieved using substitution. 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 already exists in integral table.

$∫ 1/{x.(\ln x)}\ dx = \ln (\ln x) + C$

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Last Problem

$∫ 1/{{x^2}.(\ln x)}\ dx$

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

Or —

\[∫{x^m(\ln x)^n}dx={x^{m+1}{(\ln x)^n}}/{m+1} - {n/{m+1}}∫{x^m (\ln x)^{n-1}dx}\;\;\;\; (\for \;\; m≠-1)\]

We can substitute the —

m with -2 ➡️ for x and n with -1 ➡️ for ln x

And that's the solution.

⬆️ We don't need to "solve" the second integral.

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See you next time. 👋