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

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

\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

\frac{d}{d x} (3 (cos (x))^{2})

n = 1 \sum \infty \frac{1}{2 ^{n} + 1}

\frac{d}{d x} (\frac{1}{e ^{x^{2}}})

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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