How do you integrate $\int ln x d x$ $int lnx dx$ using integration by parts?

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How do you integrate $\int ln x d x$ $int lnx dx$ using integration by parts?

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Answer 1 · 2016-01-05T16:28:08

$\int ln x d x = x ln x - x + C$ $int lnxdx = x ln x - x + C$

Explanation:

The formula for integration by parts, just for reference:

$\int u d v = u v - \int v d u$ $int u dv = uv - int v du$

It's not all that obvious what we should set $u$ $u$ and $d v$ $dv$ as, since we only have one function here.

However, let's set $u = ln x$ $u = ln x$ . Then $d u = \frac{1}{x} d x$ $du = 1/x dx$ .

The real trick is to set $d v = d x$ $dv = dx$ , making $v = x$ $v = x$ . This will allow us to get rid of the $ln$ $ln$ and simply be left with a polynomial to integrate.

Plugging into our formula we have

$\int ln x d x = x ln x - \int \frac{x}{x} d x$ $int ln x dx = x ln x - int x/x dx$

$\frac{x}{x}$ $x/x$ is just one:

$\int ln x d x = x ln x - \int d x$ $int ln x dx = x ln x - int dx$

And we know $\int d x$ $int dx$ turns out to be $x$ $x$ . Don't forget the constant of integration:

$\int ln x d x = x ln x - x + C$ $int ln x dx = x ln x - x + C$

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How do you integrate $\int ln x d x$ $int lnx dx$ using integration by parts?

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