How do you evaluate the integral of $\frac{ln (2 x)}{x^{2}} d x$ $ln(2x)/x^2 dx$ ?

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How do you evaluate the integral of $\frac{ln (2 x)}{x^{2}} d x$ $ln(2x)/x^2 dx$ ?

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Answer 1 · 2015-03-16T01:14:16

You can evaluate this integral by using integration by parts.

Let $u = ln (2 x)$ $u=ln(2x)$

Let $d v = \frac{1}{x^{2}} d x$ $dv=1/x^2dx$

Differentiating $u = ln (2 x)$ $u=ln(2x)$ we have

$d u = \frac{1}{x} d x$ $du=1/xdx$

Integrating $d v = \frac{1}{x^{2}} d x$ $dv=1/x^2dx$ we have

$\int d v = \int x^{- 2} d x$ $intdv=intx^-2dx$

$v = - \frac{1}{x}$ $v=-1/x$

Recall the integration by parts formula

$u v - \int v d u$ $uv-intvdu$

Now proceed as follows

$- \frac{ln (2 x)}{x} - \int - \frac{1}{x} (\frac{1}{x}) d x$ $-ln(2x)/x-int-1/x(1/x)dx$

$- \frac{ln (2 x)}{x} - \int - \frac{1}{x^{2}} d x$ $-ln(2x)/x-int-1/x^2dx$

$- \frac{ln (2 x)}{x} + \int x^{- 2} d x$ $-ln(2x)/x+intx^-2dx$

$- \frac{ln (2 x)}{x} - \frac{1}{x} + C$ $-ln(2x)/x-1/x+C$

We can rewrite if you like

$- \frac{(ln (2 x) + 1)}{x} + C$ $-((ln(2x)+1))/x+C$

General note:
For $n \neq - 1$ $n!=-1$ , any integral of the form $\int x^{n} ln x d x$ $intx^nlnxdx$ can be found by the process described above.

(For $n = - 1$ $n=-1$ use $u$ $u$ -substitution, with $u = ln x$ $u=lnx$ so $d u = \frac{1}{x} d x$ $du=1/xdx$ .)

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How do you evaluate the integral of $\frac{ln (2 x)}{x^{2}} d x$ $ln(2x)/x^2 dx$ ?

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