Key to "Is there a different approach to evaluate ∫ln⁡(x)dx?"

tebollahb

Answered question

2022-01-21

Is there a different approach to evaluate

\int \ln (x) d x

Melinda McCombs

Beginner2022-01-21Added 38 answers

\ln x = \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \frac{1}{n} {(x - 1)}^{n}

\int \ln x d x = \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \frac{1}{n} \cdot \frac{1}{1 + n} {(x - 1)}^{n + 1}

= \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{1}{n} - \frac{1}{n + 1}) {(x - 1)}^{n + 1}

= (x - 1) \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \frac{1}{n} {(x - 1)}^{n} + \sum_{n = 1}^{\infty} {(- 1)}^{n + 2} \frac{1}{n + 1} {(x - 1)}^{n + 1}

= (x - 1) \ln x + \ln x - (x - 1)

= x \ln x - x + C

Nadine Salcido

Beginner2022-01-22Added 34 answers

RizerMix

Expert2022-01-27Added 656 answers

For fun, a bit of trickery: Let

x = e^{y}

. Then:

\int \ln x d x = \int y e^{y} d y;

Now look at :

\frac{d}{d y} (y e^{y}) = e^{y} + y e^{y} .

Thus:

\int y e^{y} d y =

\int \frac{d}{d y} (y e^{y}) - \int e^{y} d y =

y e^{y} - e^{y} = x \ln x - x + C .

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