How do you integrate $\int \frac{ln (x)}{x} d x$ $int ln(x)/x dx$ using integration by parts?

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How do you integrate $\int \frac{ln (x)}{x} d x$ $int ln(x)/x dx$ using integration by parts?

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Answer 1 · 2015-12-20T16:34:33

$\int \frac{ln (x)}{x} d x = \frac{{ln (x)}^{2}}{4}$ $intln(x)/xdx = ln(x)^2/4$

Explanation:

Integration by parts is a bad idea here, you will constantly have $\int \frac{ln (x)}{x} d x$ $intln(x)/xdx$ somewhere. It is better to change the variable here because we know that the derivative of $ln (x)$ $ln(x)$ is $\frac{1}{x}$ $1/x$ .

We say that $u (x) = ln (x)$ $u(x) = ln(x)$ , it implies that $d u = \frac{1}{x} d x$ $du = 1/xdx$ . We now have to integrate $\int u d u$ $intudu$ .

$\int u d u = \frac{u^{2}}{2}$ $intudu = u^2/2$ so $\int \frac{ln (x)}{x} d x = \frac{{ln (x)}^{2}}{2}$ $intln(x)/xdx = ln(x)^2/2$

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How do you integrate $\int \frac{ln (x)}{x} d x$ $int ln(x)/x dx$ using integration by parts?

1 Answer

Explanation:

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How do you integrate $\int \frac{ln (x)}{x} d x$ $int ln(x)/x dx$ using integration by parts?