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Unlocking the Integral of ln(x)

An exploration of how integration by parts transforms a difficult logarithmic integral into a simple algebraic one, followed by a definite evaluation.

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The Hidden Product
∫ln⁡(x)dx=∫ln⁡(x)⋅1dx\int \ln(x) dx = \int \ln(x) \cdot 1 dx
How do we integrate a function that doesn't have an obvious anti-derivative? The trick is to realize that ln(x) can be viewed as the product of ln(x) and 1, inviting the strategy of integration by parts.
Step-by-step solver
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(a) Setup

Rewrite the integral as a product and define u and dv.

∫ln⁡x⋅1dx;u=ln⁡x,dv=dx  ⟹  du=1xdx,v=x\int \ln x \cdot 1 dx; \quad u = \ln x, dv = dx \implies du = \frac{1}{x}dx, v = x
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(a) Integration

Apply the integration by parts formula.

∫ln⁡xdx=xln⁡x−∫x⋅1xdx=xln⁡x−x+C\int \ln x dx = x\ln x - \int x \cdot \frac{1}{x} dx = x\ln x - x + C
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(b) Limits

Evaluate the result at the limits e and 1.

∫1eln⁡xdx=[xln⁡x−x]1e\int_1^e \ln x dx = [x\ln x - x]_1^e
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(b) Final Calculation

Substitute the values to find the final result.

(eln⁡e−e)−(1ln⁡1−1)=(e−e)−(0−1)=1(e \ln e - e) - (1 \ln 1 - 1) = (e - e) - (0 - 1) = 1

Original question

Use integration by parts to find the integral of ln(x) with respect to x, and hence evaluate the definite integral from 1 to e.

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