How do you find the integral $\frac{ln (x)}{x^{2}}$ $ln(x)/(x^2)$ ?

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How do you find the integral $\frac{ln (x)}{x^{2}}$ $ln(x)/(x^2)$ ?

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Answer 1 · 2015-09-07T11:12:01

$- \frac{ln x}{x} - \frac{1}{x} + C$ $-lnx /x -1/x +C$

Explanation:

Integration by parts can be done in this case, $\int \frac{ln x}{x^{2}} d x$ $int (ln x)/x^2 dx$ = $\int ln x \cdot \frac{d}{d x} (- \frac{1}{x}) d x$ $int lnx*d/dx(-1/x) dx$ = $- \frac{ln x}{x} - \int \frac{d}{d x} (ln x) \cdot \frac{- 1}{x} d x + C$ $-lnx /x -int d/dx (lnx)*(-1)/x dx+C$ = $- \frac{ln x}{x} + \int \frac{1}{x^{2}} d x + C$ $-lnx /x+int 1/x^2 dx +C$ = $- \frac{ln x}{x} - \frac{1}{x} + C$ $-lnx /x -1/x +C$

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How do you find the integral $\frac{ln (x)}{x^{2}}$ $ln(x)/(x^2)$ ?

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How do you find the integral $\frac{ln (x)}{x^{2}}$ $ln(x)/(x^2)$ ?