What is the integral of (4x^2+4x)/(2x-1) ?

The integral of (4x^2+4x)/(2x-1) is 1/4 (4x^2+12x+6ln|2x-1|-7)+C

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integral of (4x^2+4x)/(2x-1)

\int \frac{4 x ^{2} + 4 x}{2 x - 1} d x

\frac{1}{4} (4 x^{2} + 12 x + 6 ln ∣ 2 x - 1 ∣ - 7) + C

Solution steps

\int \frac{4 x ^{2} + 4 x}{2 x - 1} d x

Apply u-substitution

: \int \frac{( u + 3 ) ( u + 1 )}{2 u} d u

= \int \frac{( u + 3 ) ( u + 1 )}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{( u + 3 ) ( u + 1 )}{u} d u

Expand

\frac{( u + 3 ) ( u + 1 )}{u} : u + 4 + \frac{3}{u}

= \frac{1}{2} \cdot \int u + 4 + \frac{3}{u} 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}{2} (\int u d u + \int 4 d u + \int \frac{3}{u} d u)

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

\int 4 d u = 4 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{2} (\frac{u ^{2}}{2} + 4 u + 3 ln ∣ u ∣)

Substitute back

u = 2 x - 1

= \frac{1}{2} (\frac{( 2 x - 1 ) ^{2}}{2} + 4 (2 x - 1) + 3 ln ∣ 2 x - 1 ∣)

Simplify

\frac{1}{2} (\frac{( 2 x - 1 ) ^{2}}{2} + 4 (2 x - 1) + 3 ln ∣ 2 x - 1 ∣) : \frac{1}{4} (4 x^{2} + 12 x + 6 ln ∣ 2 x - 1 ∣ - 7)

= \frac{1}{4} (4 x^{2} + 12 x + 6 ln ∣ 2 x - 1 ∣ - 7)

Add a constant to the solution

= \frac{1}{4} (4 x^{2} + 12 x + 6 ln ∣ 2 x - 1 ∣ - 7) + C

\int \frac{1}{- x ^{2} - 4 x + 21} d x

\int \frac{2 e ^{x}}{1 - e ^{2 x}} d x

9 x^{2} sin (x) tan (x)

\int a cos (t) d t

\frac{d y}{d x} = \frac{5 ( 3 - y ) ^{2}}{( 3 - x ) ^{2}}

What is the integral of (4x^2+4x)/(2x-1) ?
The integral of (4x^2+4x)/(2x-1) is 1/4 (4x^2+12x+6ln|2x-1|-7)+C