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

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

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

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

2 x^{2} + ln ∣ x - 1 ∣ + C

Solution steps

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

Long division

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

= \int 4 x + \frac{1}{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 4 x d x + \int \frac{1}{x - 1} d x

\int 4 x d x = 2 x^{2}

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= 2 x^{2} + ln ∣ x - 1 ∣

Add a constant to the solution

= 2 x^{2} + ln ∣ x - 1 ∣ + C

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