Question

integral of( log x) power 2

Answer 1

Answer on Question #54549 – Math – Integral Calculus

Find:

$$\int (\log x) ^ {2} d x$$

Solution

$$\begin{array}{l} \int (\log x) ^ {2} d x = \int \log^ {2} x d x = \text{(integration by parts } u = \log^ {2} x, dv = dx) \\ = x \log^ {2} x - \int x d (\log^ {2} x) = x \log^ {2} x - \int x * 2 \log x * \frac {1}{x} d x = \\ = x \log^ {2} x - 2 \int \log x d x = \\ = \text{(integration by parts } u = \log x, dv = dx) = \\ = x \log^ {2} x - 2 \left(x \log x - \int x d (\log x)\right) = \\ = x \log^ {2} x - 2 x \log x + 2 \int x d (\log x) = \\ = x \log^ {2} x - 2 x \log x + 2 \int x * \frac {1}{x} d x = x \log^ {2} x - 2 x \log x + 2 \int 1 d x = \\ = x \log^ {2} x - 2 x \log x + 2 x + c = x (\log^ {2} x - 2 \log x + 2) + c, \end{array}$$

where $c$ is an integration constant.

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