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

The integral of (x^2)/((2x-1)^2) is (2x^2+2xln|2x-1|-2x-ln|2x-1|)/(4(2x-1))+C

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

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

\frac{2 x ^{2} + 2 x ln ∣ 2 x - 1 ∣ - 2 x - ln ∣ 2 x - 1 ∣}{4 ( 2 x - 1 )} + C

Solution steps

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

Apply u-substitution

= \int \frac{( u + 1 ) ^{2}}{8 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{( u + 1 ) ^{2}}{u ^{2}} d u

Expand

\frac{( u + 1 ) ^{2}}{u ^{2}} : 1 + \frac{2}{u} + \frac{1}{u ^{2}}

= \frac{1}{8} \cdot \int 1 + \frac{2}{u} + \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int 1 d u + \int \frac{2}{u} d u + \int \frac{1}{u ^{2}} d u)

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= \frac{1}{8} (u + 2 ln ∣ u ∣ - \frac{1}{u})

Substitute back

u = 2 x - 1

= \frac{1}{8} (2 x - 1 + 2 ln ∣ 2 x - 1 ∣ - \frac{1}{2 x - 1})

Simplify

\frac{1}{8} (2 x - 1 + 2 ln ∣ 2 x - 1 ∣ - \frac{1}{2 x - 1}) : \frac{2 x ^{2} + 2 x ln ∣ 2 x - 1 ∣ - 2 x - ln ∣ 2 x - 1 ∣}{4 ( 2 x - 1 )}

= \frac{2 x ^{2} + 2 x ln ∣ 2 x - 1 ∣ - 2 x - ln ∣ 2 x - 1 ∣}{4 ( 2 x - 1 )}

Add a constant to the solution

= \frac{2 x ^{2} + 2 x ln ∣ 2 x - 1 ∣ - 2 x - ln ∣ 2 x - 1 ∣}{4 ( 2 x - 1 )} + C

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What is the integral of (x^2)/((2x-1)^2) ?
The integral of (x^2)/((2x-1)^2) is (2x^2+2xln|2x-1|-2x-ln|2x-1|)/(4(2x-1))+C