What is the integral of (6x^2)/((2x-1)(2x+3)) ?

The integral of (6x^2)/((2x-1)(2x+3)) is 6(1/4 x+1/32 (-4ln|4x^2+4x-3|-5ln|x+3/2 |+5ln|x-1/2 |))+C

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integral of (6x^2)/((2x-1)(2x+3))

\int \frac{6 x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} d x

6 (\frac{1}{4} x + \frac{1}{32} (- 4 ln 4 x^{2} + 4 x - 3 - 5 ln x + \frac{3}{2} + 5 ln x - \frac{1}{2})) + C

Solution steps

\int \frac{6 x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} d x

Long division

\frac{x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} : \frac{1}{4} + \frac{- 4 x + 3}{4 ( 4 x ^{2} + 4 x - 3 )}

= 6 \cdot \int \frac{1}{4} + \frac{- 4 x + 3}{4 ( 4 x ^{2} + 4 x - 3 )} 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

= 6 (\int \frac{1}{4} d x + \int \frac{- 4 x + 3}{4 ( 4 x ^{2} + 4 x - 3 )} d x)

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

\int \frac{- 4 x + 3}{4 ( 4 x ^{2} + 4 x - 3 )} d x = \frac{1}{32} (- 4 ln 4 x^{2} + 4 x - 3 - 5 ln x + \frac{3}{2} + 5 ln x - \frac{1}{2})

= 6 (\frac{1}{4} x + \frac{1}{32} (- 4 ln 4 x^{2} + 4 x - 3 - 5 ln x + \frac{3}{2} + 5 ln x - \frac{1}{2}))

Add a constant to the solution

= 6 (\frac{1}{4} x + \frac{1}{32} (- 4 ln 4 x^{2} + 4 x - 3 - 5 ln x + \frac{3}{2} + 5 ln x - \frac{1}{2})) + C

(x^{2} - 3 x y) d x + x d y = 0

\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} - 5 y = 4

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

x (2 x + 1)^{4}

\frac{d}{d x} (e^{x} cos (y))

What is the integral of (6x^2)/((2x-1)(2x+3)) ?
The integral of (6x^2)/((2x-1)(2x+3)) is 6(1/4 x+1/32 (-4ln|4x^2+4x-3|-5ln|x+3/2 |+5ln|x-1/2 |))+C