What is the integral of 2/(x(3x-8)) ?

The integral of 2/(x(3x-8)) is 2(-1/8 ln|x|+1/8 ln|3x-8|)+C

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integral of 2/(x(3x-8))

\int \frac{2}{x ( 3 x - 8 )} d x

2 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ 3 x - 8 ∣) + C

Solution steps

\int \frac{2}{x ( 3 x - 8 )} 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}{x ( 3 x - 8 )} d x

Take the partial fraction of

\frac{1}{x ( 3 x - 8 )} : - \frac{1}{8 x} + \frac{3}{8 ( 3 x - 8 )}

= 2 \cdot \int - \frac{1}{8 x} + \frac{3}{8 ( 3 x - 8 )} 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{1}{8 x} d x + \int \frac{3}{8 ( 3 x - 8 )} d x)

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

\int \frac{3}{8 ( 3 x - 8 )} d x = \frac{1}{8} ln ∣ 3 x - 8 ∣

= 2 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ 3 x - 8 ∣)

Add a constant to the solution

= 2 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ 3 x - 8 ∣) + C

\frac{d}{d x} (x in (x))

\int ln (4 + x^{2}) d x

\int (2 x + y) d x

f (x) = arctan (x^{4})

\int \frac{3}{x ^{2} + 1} d x

What is the integral of 2/(x(3x-8)) ?
The integral of 2/(x(3x-8)) is 2(-1/8 ln|x|+1/8 ln|3x-8|)+C