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

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

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

integral

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

\frac{( x - 1 ) ^{2}}{x} - 2 (ln ∣ x ∣ + \frac{1}{x}) + C

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply Integration By Parts

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

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

\int \frac{2 ( x - 1 )}{x ^{2}} d x = 2 (ln ∣ x ∣ + \frac{1}{x})

= \frac{( x - 1 ) ^{2}}{x} - 2 (ln ∣ x ∣ + \frac{1}{x})

Add a constant to the solution

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

t^{3} + 3 t^{2} - 2 t + t^{- 1}

\frac{1}{tan ( x )}

\frac{x}{81 x ^{4} - 1}

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

\frac{1}{( 4 - x ^{2} )}

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