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

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

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

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

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

Solution steps

\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

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\frac{d}{d x} (\frac{4}{3} x^{3} - 4 x + 3)

f (x) = \frac{1}{x - 4}

\int_{0}^{\frac{π}{3}} sin (x) ln (cos (x)) d x

\int cos^{2} (ln (x)) d x

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