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

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

Popular >

integral of (1/x-1)^2

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

- \frac{1}{x} - 2 ln ∣ x ∣ + x + C

Solution steps

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

Expand

(\frac{1}{x} - 1)^{2} : \frac{1}{x ^{2}} - \frac{2}{x} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} - 2 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{1}{x} - 2 ln ∣ x ∣ + x + C

f (x) = cos (\frac{x + 1}{x - 1})

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

x \to 0 lim (x^{3} cos (\frac{2}{x}))

\int \frac{30}{( x - 1 ) ( x ^{2} + 9 )} d x

f (x) = \frac{4 x ^{3}}{x ^{4} + 1}

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