What is the integral of 2/((8x-2)(4x-6)(11x-10)) ?

The integral of 2/((8x-2)(4x-6)(11x-10)) is 2(1/290 ln|4x-1|+1/260 ln|2x-3|-11/1508 ln|11x-10|)+C

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integral of 2/((8x-2)(4x-6)(11x-10))

\int \frac{2}{( 8 x - 2 ) ( 4 x - 6 ) ( 11 x - 10 )} d x

2 (\frac{1}{290} ln ∣ 4 x - 1 ∣ + \frac{1}{260} ln ∣ 2 x - 3 ∣ - \frac{11}{1508} ln ∣ 11 x - 10 ∣) + C

Solution steps

\int \frac{2}{( 8 x - 2 ) ( 4 x - 6 ) ( 11 x - 10 )} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{( 8 x - 2 ) ( 4 x - 6 ) ( 11 x - 10 )} d x

Take the partial fraction of

\frac{1}{( 8 x - 2 ) ( 4 x - 6 ) ( 11 x - 10 )} : \frac{2}{145 ( 4 x - 1 )} + \frac{1}{130 ( 2 x - 3 )} - \frac{121}{1508 ( 11 x - 10 )}

= 2 \cdot \int \frac{2}{145 ( 4 x - 1 )} + \frac{1}{130 ( 2 x - 3 )} - \frac{121}{1508 ( 11 x - 10 )} 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

= 2 (\int \frac{2}{145 ( 4 x - 1 )} d x + \int \frac{1}{130 ( 2 x - 3 )} d x - \int \frac{121}{1508 ( 11 x - 10 )} d x)

\int \frac{2}{145 ( 4 x - 1 )} d x = \frac{1}{290} ln ∣ 4 x - 1 ∣

\int \frac{1}{130 ( 2 x - 3 )} d x = \frac{1}{260} ln ∣ 2 x - 3 ∣

\int \frac{121}{1508 ( 11 x - 10 )} d x = \frac{11}{1508} ln ∣ 11 x - 10 ∣

= 2 (\frac{1}{290} ln ∣ 4 x - 1 ∣ + \frac{1}{260} ln ∣ 2 x - 3 ∣ - \frac{11}{1508} ln ∣ 11 x - 10 ∣)

Add a constant to the solution

= 2 (\frac{1}{290} ln ∣ 4 x - 1 ∣ + \frac{1}{260} ln ∣ 2 x - 3 ∣ - \frac{11}{1508} ln ∣ 11 x - 10 ∣) + C

x \to - 14.5 lim (ln (x^{2} - 9))

\int \frac{2 x}{x ^{2} + 3} d x

x \to 0 lim (x^{x^{4}})

\frac{d x}{d t} = 0.8 x - 0.004 x^{2}

x \to 0 lim (\frac{( 1 + x ) ^{n} - 1}{x})

What is the integral of 2/((8x-2)(4x-6)(11x-10)) ?
The integral of 2/((8x-2)(4x-6)(11x-10)) is 2(1/290 ln|4x-1|+1/260 ln|2x-3|-11/1508 ln|11x-10|)+C