- #1

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so is there an integral of ln x?

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- Thread starter lsim16
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- #1

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so is there an integral of ln x?

- #2

Tide

Science Advisor

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Have you learned about integrating by parts?

- #3

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this is only a maths b class

does this mean, there is no simple way of doing this?

- #4

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[tex] \int u dv = uv - \int v du [/tex]

- #5

TD

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It should be [itex] \int ln x dx = x \ln x - x = x(\ln x - 1) [/itex]whozum said:

[tex] \int u dv = uv - \int v du [/tex]

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- #7

Curious3141

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The other way is to make the obvious substitution [itex]x = e^u[/tex], then integrate by parts.

- #8

Curious3141

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Let's say you know that [tex]\int f(x)dx = g(x) + c[/tex]. You want to find [tex]\int f^{-1}(x)dx[/tex] where the exponent denotes the inverse function.

Substitute [tex]x = f(u)[/tex].

You get [tex]\int f^{-1}(x)dx = \int uf'(u)du = uf(u) - \int f(u)du = uf(u) - g(u) + c = xf^{-1}(x) - g(f^{-1}(x)) + c[/tex]

You can verify that works by trying out [tex]f^{-1}(x) = \ln x[/tex] in the result.

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