What is the integral of (12x^2)/(2x+1) ?

The integral of (12x^2)/(2x+1) is 3/4 (4x^2-4x+2ln|2x+1|-3)+C

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integral of (12x^2)/(2x+1)

\int \frac{12 x ^{2}}{2 x + 1} d x

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

Solution steps

\int \frac{12 x ^{2}}{2 x + 1} d x

Take the constant out:

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

= 12 \cdot \int \frac{x ^{2}}{2 x + 1} d x

Apply u-substitution

: \int \frac{( u - 1 ) ^{2}}{8 u} d u

= 12 \cdot \int \frac{( u - 1 ) ^{2}}{8 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

Expand

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

= 12 \cdot \frac{1}{8} \cdot \int u - 2 + \frac{1}{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

= 12 \cdot \frac{1}{8} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 12 \cdot \frac{1}{8} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Substitute back

u = 2 x + 1

= 12 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣)

Simplify

12 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣) : \frac{3}{4} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

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

Add a constant to the solution

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

x \to 0 lim (\frac{4 x ^{2} - 2 x}{x})

\frac{\partial}{\partial x} (ln (x^{2} + 3 y^{2} + 8))

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

\frac{d}{d x} (a x^{4} - 2 a x^{3} + 2 b x^{2} \cdot ln (\frac{x}{2}))

x + 2

What is the integral of (12x^2)/(2x+1) ?
The integral of (12x^2)/(2x+1) is 3/4 (4x^2-4x+2ln|2x+1|-3)+C