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)

integral

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

\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

\frac{t}{6}

\frac{1}{( 1 + x ^{3} )}

(3 x - 5)^{- 2}

\frac{x}{y ^{2}}

r^{2} - x^{2}

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