What is the integral of 1/(x^2)x' ?

The integral of 1/(x^2)x' is (x')/x-2/x+C

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integral of 1/(x^2)x'

\int \frac{1}{x ^{2}} x^{'} d x

\frac{x '}{x} - \frac{2}{x} + C

Solution steps

\int \frac{1}{x ^{2}} x^{'} d x

Apply Integration By Parts

: \frac{x ^{'}}{x} - \int - \frac{2}{x ^{2}} d x

= \frac{x ^{^{'}}}{x} - \int - \frac{2}{x ^{2}} d x

\int - \frac{2}{x ^{2}} d x = \frac{2}{x}

= \frac{x ^{^{'}}}{x} - \frac{2}{x}

Add a constant to the solution

= \frac{x ^{^{'}}}{x} - \frac{2}{x} + C

(5 x - 3) y^{^{''}} + y^{^{'}} = 0

6

\frac{1}{x}, x = 2

x \to - 6 + lim (\frac{- 9 x}{x + 6})