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# What is the integral of (lnx)/x?

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May 26, 2018

$\int \ln \frac{x}{x} \mathrm{dx} = {\ln}^{2} \frac{x}{2} + C$

#### Explanation:

You can think of Integration by Substitution like a game where the goal is to take the derivative of one factor of the integrand and use the result of taking that derivative to cancel out the other factor.

For instance, the derivative of $\ln x$ is $\frac{1}{x}$ which looks promising since it's the same as the other factor.

Let $u = \ln x$. Taking the derivative of both sides:

$\frac{d}{\mathrm{dx}}$ u = $\frac{d}{\mathrm{dx}}$ $\ln x$

$\frac{\mathrm{du}}{\mathrm{dx}}$ = $\frac{1}{x}$

$\mathrm{dx}$ = $x \cdot \mathrm{du}$

If we substitute our findings above into the original integral, we see that:

$\int \ln \frac{x}{x} \mathrm{dx} = \int \frac{u}{x} \cdot x \cdot \mathrm{du}$

We see that on the right side of the equation above, the x cancels yielding:

$\int \frac{u}{\cancel{x}} \cdot \cancel{x} \cdot \mathrm{du} = \int \left(u\right) \mathrm{du}$

Therein lies the game: we cancelled out a factor in order to get something we know how to integrate:

$\int \left(u\right) \mathrm{du} = {u}^{2} / 2 + C$

The original integrand was in x, so we'll provide the solution in x by substituting back:

${u}^{2} / 2 + C = {\ln}^{2} \frac{x}{2} + C$

Therefore,

$\int \ln \frac{x}{x} \mathrm{dx} = {\ln}^{2} \frac{x}{2} + C$

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Jim S Share
May 26, 2018

$\int \ln \frac{x}{x} \mathrm{dx} = {\ln}^{2} \frac{x}{2} + c$

#### Explanation:

$\int \ln \frac{x}{x} \mathrm{dx}$

Substitute

$\ln x = u$

$\frac{1}{x} \mathrm{dx} = \mathrm{du}$

$=$ $\int u \mathrm{du} = {u}^{2} / 2 + c = {\ln}^{2} \frac{x}{2} + c$

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sjc Share
May 26, 2018

$\frac{1}{2} {\left(\ln x\right)}^{2} + {c}_{2}$

#### Explanation:

we need integration by parts
$I = \int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

$\int \ln \frac{x}{x} \mathrm{dx}$

$u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x} \implies v = \ln x$

$I = {\left(\ln x\right)}^{2} - \int \ln \frac{x}{x} \mathrm{dx} + {c}_{1}$

$I = {\left(\ln x\right)}^{2} - I + {c}_{1}$

$2 I = {\left(\ln x\right)}^{2} + {c}_{1}$

$I = \frac{1}{2} {\left(\ln x\right)}^{2} + {c}_{2}$

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