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

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

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integral of (1-2x)/(x^2)

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

- \frac{1 - 2 x}{x} - 2 ln ∣ x ∣ + C

Solution steps

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

Apply Integration By Parts

: - \frac{1 - 2 x}{x} - \int \frac{2}{x} d x

= - \frac{1 - 2 x}{x} - \int \frac{2}{x} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= - \frac{1 - 2 x}{x} - 2 ln ∣ x ∣

Add a constant to the solution

= - \frac{1 - 2 x}{x} - 2 ln ∣ x ∣ + C

\frac{d}{d x} (ln (3 x - 2))

f (x) = cos (2 x), at x = \frac{π}{6}

\frac{d}{d t} (t^{2} - 2)

\frac{\partial}{\partial z} (4 x - yz + 3 y z^{2} + z^{2})

What is the integral of (1-2x)/(x^2) ?
The integral of (1-2x)/(x^2) is -(1-2x)/x-2ln|x|+C